Brilliant Summation Contest Season-1 (Part 2)

We have a proof on brilliant

$\sum _{ n=-\infty }^{ \infty }{ { \left( e^{-\pi} \right) }^{ { n }^{ 2 } } } =\frac { { \left( \pi \right) }^{ \frac { 1 }{ 4 } } }{ \Gamma \left( \frac { 3 }{ 4 } \right) }$

Now use this formula $\phi(-e^{-2\pi/n})=\frac { { \left( \pi \right) }^{ \frac { 1 }{ 4 } } }{ \Gamma \left( \frac { 3 }{ 4 } \right) } \frac{n^{1/4}r_{2,2/n^2}}{2^{1/4}r_{2,2}}$

$r_{k,n}$ is a function of DedeKind Eta function. Which have some formulaes to solve.

$r_{2,2}=2^{1/8} \text{and} r_{2,1/2}=2^{-1/8}$

Thus , you get $\frac{(\frac{\pi}{2})^{1/4}}{\Gamma(3/4)}$

Solution to Problem 17:

$S=\sum _{ n=0 }^{ \infty } \frac { 2^{ n-1 }n! }{ \prod _{ r=0 }^{ n } (4r+1) }$

$S=\sum _{ n=0 }^{ \infty } \left[ \frac { { 2 }^{ -n-1 }\Gamma \left( n+1 \right) \Gamma \left( \frac { 5 }{ 4 } \right) }{ \Gamma \left( n+\frac { 5 }{ 4 } \right) } \right]$

Using beta function, we get:

$S=\sum _{ n=0 }^{ \infty } \left[ { 2 }^{ -n-1 }B\left( n+1,\frac { 5 }{ 4 } \right) .\left( n+\frac { 5 }{ 4 } \right) \right]$

$S=\frac { 1 }{ 2 } \sum _{ n=0 }^{ \infty } \left[ { 2 }^{ -n }B\left( n+1,\frac { 5 }{ 4 } \right) .n \right] +\frac { 5 }{ 8 } \sum _{ n=0 }^{ \infty } \left[ { 2 }^{ -n }B\left( n+1,\frac { 5 }{ 4 } \right) \right]$

$S=\frac { 1 }{ 2 } \int _{ 0 }^{ 1 }{ \left\{ \sum _{ n=0 }^{ \infty }{ \left[ { 2 }^{ -n }.n.{ t }^{ n }{ \left( 1-t \right) }^{ \frac { 1 }{ 4 } } \right] } dt \right\} } +\frac { 5 }{ 8 } \int _{ 0 }^{ 1 }{ \left\{ \sum _{ n=0 }^{ \infty }{ \left[ { 2 }^{ -n }.{ t }^{ n }{ \left( 1-t \right) }^{ \frac { 1 }{ 4 } } \right] } dt \right\} }$

Now I'll use the following geometric progression summations:

$A=\sum _{ n=0 }^{ \infty }{ \left[ { \left( xt \right) }^{ n }{ \left( 1-t \right) }^{ \frac { 1 }{ 4 } } \right] } =\frac { { \left( 1-t \right) }^{ \frac { 1 }{ 4 } } }{ 1-xt } ,\quad B=\sum _{ n=0 }^{ \infty }{ \left[ { \left( xt \right) }^{ n }{ .n.\left( 1-t \right) }^{ \frac { 1 }{ 4 } } \right] } =\frac { { xt\left( 1-t \right) }^{ \frac { 1 }{ 4 } } }{ { \left( 1-xt \right) }^{ 2 } }$

$S=\frac { 5 }{ 4 } \int _{ 0 }^{ 1 }{ \frac { { \left( 1-t \right) }^{ \frac { 1 }{ 4 } } }{ 2-t } dt } +\int _{ 0 }^{ 1 }{ \frac { x{ \left( 1-x \right) }^{ \frac { 1 }{ 4 } } }{ { \left( 2-x \right) }^{ 2 } } dx }$

The first integral can be easily evaluated. The second one is also easily evaluated. I was screwed up and I couldn't evaluate it. So I asked for it here.

Hence the final answer on doing some simplification is:

$\boxed{\displaystyle\sum_{n=0}^{\infty} \dfrac{2^{n-1} n!}{\prod_{r=0}^{n} (4r+1)} = \dfrac{\pi + 2\log(1+\sqrt{2})}{4 \sqrt{2}}}$

Solution Of Problem 18 :

Proposition : $\text{S}_{n} = \sum_{r=0}^{n} (-1)^r \dbinom{2n+1-r}{r} = \dfrac{2}{\sqrt{3}} \sin \left( \dfrac{2 (n+1) \pi}{3} \right)$

Proof : Since $\dbinom{n}{r} = 0$ for $r > n$, we can rewrite the sum as

$\text{S}_{n} = \sum_{r=0}^{\infty} (-1)^r \dbinom{2n+1-r}{r}$

From the Binomial Theorem, we see that the sum is the coefficient of $x^n$ in

$f(x) = x^n (1-x)^{2n+1} + x^{n-1} (1-x)^{2n} + \ldots$

$= \dfrac{(1-x)^{n+1}}{x^2-x+1}$

Note that,

$\sum_{n=0}^{\infty}{{U}_{r}(a) {x}^{r}} = \dfrac{1}{x^2 -2ax+1}$

where $U_{r} (x)$ is the Chebyshev Polynomial of the second kind.

Putting $a = \dfrac{1}{2}$, we see that,

$f(x) = (1-x)^{n+1} \sum_{n=0}^{\infty}{{U}_{r} \left( \dfrac{1}{2} \right) {x}^{r}} \quad (*)$

To calculate the coefficient of $x^n$ in $(*)$, we see that coefficient of $x^{n-r}$ for fixed $r$ in $(1-x)^{n+1}$ is $(-1)^{n-r} \dbinom{n+1}{n-r}$, so coefficient of $x^n$ is,

$\sum_{r=0}^{n} (-1)^{n-r} \dbinom{n+1}{n-r} U_{r} \left(\dfrac{1}{2}\right)$

Substituting $r \mapsto r-1$, we have,

$\text{S}_{n} = \sum_{r=1}^{n+1} (-1)^{n+1-r} \dbinom{n+1}{n+1-r} U_{r-1} \left(\dfrac{1}{2}\right)$

$= (-1)^{n+1} \sum_{r=1}^{n+1} (-1)^{r} \dbinom{n+1}{r} U_{r-1} \left(\dfrac{1}{2}\right) \quad \quad \left( \because \dbinom{n}{n-r} = \dbinom{n}{r} \right)$

$= (-1)^{n+1} \left( \dfrac{2}{\sqrt{3}} \right) \sum_{r=1}^{n+1} (-1)^{r} \dbinom{n+1}{r} \sin \left( \dfrac{r \pi}{3} \right) \quad \quad \left( \because U_{r-1} \left(\dfrac{1}{2}\right) = \sin \left( \dfrac{r \pi}{3} \right) \right)$

$(-1)^{n+1} \left( \dfrac{2}{\sqrt{3}} \right) \sum_{r=0}^{n+1} (-1)^{r} \dbinom{n+1}{r} \sin \left( \dfrac{r \pi}{3} \right) \quad \quad \left( \because \sin 0 = 0 \right)$

Now, $\displaystyle \sum_{r=0}^{n+1} (-1)^{r} \dbinom{n+1}{r} \sin \left( \dfrac{r \pi}{3} \right) = \Im \left( \sum_{r=0}^{n+1} (-1)^{r} \dbinom{n+1}{r} e^{\frac{i r \pi}{3}} \right) = (-1)^{n+1} \sin \left( \dfrac{2 (n+1) \pi}{3} \right)$

$\implies \text{S}_{n} = \dfrac{2}{\sqrt{3}} \sin \left( \dfrac{2 (n+1) \pi}{3} \right) \quad \square$

Corollary : $\text{S}_{n} = \sum_{k=0}^n (-1)^{n+k} \dbinom{n+1+k}{2k+1} = \dfrac{2}{\sqrt{3}} \sin \left( \dfrac{2 (n+1) \pi}{3} \right)$

Proof : $\sum_{k=0}^n(-1)^{n+k} \dbinom{n+1+k}{2k+1} = \sum_{k=0}^n(-1)^{n+k} \dbinom{n+1+k}{n-k} = \sum_{k=0}^n(-1)^k \dbinom{2n+1-k}{k} = \text{S}_{n}$

$\left( \because \dbinom{n}{r} = \dbinom{n}{n-r} \ \text{\&} \ \sum_{r=0}^{n} f(k) = \sum_{r=0}^{n} f(n-k) \right) \quad \square$

Now,

$\lambda = \sum_{k=0}^n\dfrac{(-1)^{n+k}}{2k+1} \dbinom{n+k}{n-k}$

$= \sum_{k=0}^n \dfrac{(-1)^{n+k}(n+k)!}{(2k+1)!(n-k)!}$

$=\dfrac{1}{2n+1}\sum_{k=0}^n\dfrac{(-1)^{n+k}(n+k)!\left[(n+k+1)+(n-k)\right]}{(2k+1)!(n-k)!}$

$=\dfrac{1}{2n+1}\sum_{k=0}^n\left((-1)^{n+k} \dbinom{n+1+k}{2k+1} - (-1)^{n-1+k} \dbinom{n+k}{2k+1} \right)$

$=\dfrac{1}{2n+1} \left(\text{S}_{n} - \text{S}_{n-1}\right)$

Using the Proposition and simplifying, we have,

$\lambda = -\dfrac{2}{2n+1} \, \cos\left(\frac{2(n-1)\pi}{3}\right) \quad \square$

Note : Coefficient of $x^n$ in the Proposition can also be found by splitting $\dfrac{1}{x^2-x+1}$ as partial fraction and expanding using Infinite GP.

Let $\displaystyle S(r)=\sum_{k=0}^{n} \frac{H_{r-k}}{k}$ for $r>0$ with $S(0)=0$.

We have , $\displaystyle S(n+1) = \sum_{k=1}^{n} \frac{H_k}{n-k+1} = \sum_{k=1}^{n}\frac{H_{n-k+1}}{k} = \sum_{k=1}^{n}\frac{H_{n-k}}{k} + \frac{2H_n}{n+1} = S(n)+\frac{2H_n}{n+1}$

Telescoping above we have , $\displaystyle S(n+1) = \sum_{k=0}^{n}\frac{2H_k}{k+1}$

$\displaystyle \text{Lemma :} \large \boxed{\sum_{k=0}^{n} [(k+1)^p-k^p]H_{k}^{(m)} = (n+1)^p H_{n}^{(m)}-H_{n}^{(m-p)}}$

$\text{Proof :}$ Using summation by parts we have ,

$\displaystyle \sum_{k=0}^{n} [(k+1)^p-k^p]H_{k}^{(m)} = (n+1)^p H_{n+1}^{(p)} - \sum_{k=0}^{n} (k+1)^{(m-p)} = (n+1)^p H_{n+1}^{(p)} - H_{n+1}^{(m-p)} = (n+1)H_{n}^{(p)} - H_{n}^{(m-p)}$ & thus proved.

We'll use a well known Harmonic sum identity , $\displaystyle \sum_{k=1}^{n} \frac{H_k}{k} = \frac{1}{2}((H_n)^2+H_{n}^{(2)})$

Applying the lemma for $p=-1,m=1$ we have,

$\displaystyle S(n+1) = 2\sum_{k=0}^{n} \frac{H_k}{k+1} = \sum_{k=1}^{n} \frac{2H_k}{k} + \frac{H_n}{n+1}-H_{n}^{(2)}$

Simplifying we get , $\displaystyle \sum_{k=1}^{n} \frac{H_k}{n-k+1} = (H_{n+1})^2-H_{n+1}^{(2)}$

We have,

$\text{S} = \sum_{k=1}^{n}\dfrac{H_{k}}{n-k+1}$

$= \sum_{k=1}^{n} \sum_{r=1}^{k} \dfrac{1}{r(n-k+1)}$

$= \sum_{r=1}^{n} \sum_{k=r}^{n} \dfrac{1}{r(n-k+1)}$

$= \sum_{r=1}^{n} \sum_{k=r}^{n} \dfrac{1}{r(k-r+1)}$

$= \sum_{r=1}^{n} \sum_{k=r}^{n} \dfrac{(k-r+1 + r)}{r(k-r+1)(k+1)}$

$= \sum_{r=1}^{n} \sum_{k=r}^{n} \dfrac{1}{(k+1)} \left[\dfrac{1}{(k-r+1)} + \dfrac{1}{r}\right]$

$= \sum_{k=1}^{n} \sum_{r=1}^{k} \dfrac{1}{(k+1)} \left[\dfrac{1}{(k-r+1)} + \dfrac{1}{r}\right]$

$= 2\sum_{k=1}^{n} \dfrac{H_{k}}{k+1}$

Substitute $k+1 \mapsto k$

$\implies \text{S} = 2\sum_{k=2}^{n+1} \dfrac{H_{k-1}}{k}$

$= 2\sum_{k=2}^{n+1} \dfrac{H_{k-1} - \dfrac{1}{k}}{k}$

$= 2\sum_{k=1}^{n+1} \dfrac{H_{k}}{k} - 2H_{n+1}^{(2)}$

Since $\displaystyle \sum_{k=1}^{n+1} \dfrac{H_{k}}{k} = \dfrac{1}{2} (H_{n+1}^2 + H_{n+1}^{(2)})$ (I have proved it here), we have,

$\text{S} = (H_{n+1})^2 - H_{n+1}^{(2)} \quad \square$

Solution of Problem 24:

Lemma 1: ${ x }^{ 2 }=\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ \left[ \frac { { 2 }^{ 2n } }{ { n }^{ 2 } \dbinom{2n}{n} } \sin ^{ 2n }{ x } \right] }$

Proof:

Consider: ${ f }_{ n }\left( x \right) =\frac { \sin ^{ 2n }{ x } }{ \left( 2n \right) ! }$.

Then we get: $\\ { f" }_{ n }\left( x \right) ={ f }_{ n-1 }\left( x \right) -{ \left( 2n \right) }^{ 2 }{ f }_{ n }\left( x \right)$

We can write $\displaystyle { x }^{ 2 }=\sum _{ n=0 }^{ \infty }{ { a }_{ n }{ f }_{ n }\left( x \right) }$ for some function $a_n$.

On differentiating both sides twice:

$2=\sum _{ n=1 }^{ \infty }{ { a }_{ n }\left( { f }_{ n-1 }\left( x \right) -{ \left( 2n \right) }^{ 2 }{ f }_{ n }\left( x \right) \right) } =\sum _{ n=0 }^{ \infty }{ { a }_{ n+1 }{ f }_{ n }\left( x \right) } -\sum _{ n=1 }^{ \infty }{ { a }_{ n }{ \left( 2n \right) }^{ 2 }{ f }_{ n }\left( x \right) }$

From this relation $a_1=2$ and $a_{n+1}=a_n{(2n)}^{2}$

Thus, $\displaystyle \frac { { a }_{ n } }{ 2 } =\prod _{ k=1 }^{ n-1 }{ \frac { { a }_{ k+1 } }{ { a }_{ k } } } ={ 2 }^{ 2n-2 }{ \left( n-1 \right) ! }^{ 2 }$.

Hence, we get : $\displaystyle { x }^{ 2 }=\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ \left[ \frac { { 2 }^{ 2n } }{ { n }^{ 2 } \dbinom{2n}{n} } \sin ^{ 2n }{ x } \right] }$

Lemma 2: ${ x }^{ 4 }=\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ \left[ \frac { { 2 }^{ 2n }\sin ^{ 2n }{ x } { H }_{ n-1 }^{ (2) } }{ { n }^{ 2 }\dbinom{2n}{n} } \right] }$

Proof:

From Lemma 1, we can write: $\displaystyle { x }^{ 2 }=\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ \left[ \frac { { 2 }^{ 2n } }{ { n }^{ 2 } \dbinom{2n}{n} } \sin ^{ 2n }{ x } \right] }=\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ { b }_{ n }^{ 2 }{ f }_{ n }\left( x \right) }$

Here, $b_n={ 2 }^{ n }(n-1)!$ and $b_{n+1}=2nb_n$.

We can write: $\displaystyle { x }^{ 4 }=\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ { a }_{ n }{ b }_{ n }^{ 2 }{ f }_{ n }\left( x \right) }$ for some function $a_n$.

On differentiating both sides twice:

$12{ x }^{ 2 }=\frac { 1 }{ 2 } \sum _{ n=0 }^{ \infty }{ { a }_{ n+1 }{ b }_{ n+1 }^{ 2 }{ f }_{ n }\left( x \right) } -\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ { a }_{ n }{ b }_{ n }^{ 2 }{ \left( 2n \right) }^{ 2 }{ f }_{ n }\left( x \right) }$

Thus, we get: $({ a }_{ n+1 }-{ a }_{ n }){ b }_{ n }^{ 2 }{ \left( 2n \right) }^{ 2 }=12{ b }_{ n }^{ 2 }$

From here we get: $({ a }_{ n+1 }-{ a }_{ n })=\frac { 12 }{ { \left( 2n \right) }^{ 2 } }$.

Hence, ${ a }_{ n }={ H }_{ n-1 }^{ (2) }$.

Therefore, we get: $\displaystyle { x }^{ 4 }=\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ \left[ \frac { { 2 }^{ 2n }\sin ^{ 2n }{ x } { H }_{ n-1 }^{ (2) } }{ { n }^{ 2 }\dbinom{2n}{n} } \right] }$.

From lemma 2, we get: $\displaystyle (\sin^{-1} x)^4 = \frac32 \sum_{n=1}^\infty \frac{2^{2n} H_{n-1}^{(2)}}{n^2 \binom{2n}{n}} \,x^{2n} \ \quad ...(A)$

Now, in the problem, we use: ${ F }_{ 2k }=\frac { { \varphi }^{ 2k }-\frac { 1 }{ { \varphi }^{ 2k } } }{ \sqrt { 5 } }$

On substituting and using $A$, we get: $\boxed{\displaystyle \sum_{k \geq 1} \frac{F_{2k}H_{k-1}^{(2)}}{k^2\binom{2k}{k}}=\frac{2\pi^4}{375\sqrt5}}$

Although @Aditya Sharma has solved the question - I thought of an approach that might be considered more "illuminating" than induction.

Let's think of the $i_1, i_2,...i_m$ as numbers on a register - each $i_k$ is one number from $1$ to $n$. Initially, the register reads $m$ ones, because all the $i_k$ are $1$.
For each unique number on the register, we add one to our count.

Now, we turn the first number ($i_m$) to $2$.
The register reads $2$ followed by $m - 1$ $1$s.
Note that a digit to the right of another digit can never be greater than it, because $i_j \geq i_k$if $j \geq k$.

So, the number on our register is an $m$ digit number whose digits are non-decreasing when read from left to right.
We can keep turning the register and counting how many unique numbers we get, until we reach $m$ digits with value $n$. After this, our register stops working.

It is easy to see that every single number satisfying the non-decreasing digit condition is encountered, and only once.

So our question now becomes:
What is the number of tuples $i_1, i_2,...i_m$ satisfying,

$1 \leq i_1 \leq i_2 \leq \ldots \leq i_m \leq n$

Because for each one of these tuples, we have a number on our register.

If we write the numbers in a row, $i_1 i_2 \ldots i_m$ and divide them into $n$ blocks, such that the numbers enclosed in the $n^{th}$ block get value $n$, then we have found one such possibility. We need $n - 1$ bars to separate the numbers into $n$ blocks.
For example, if $m = 4, n = 4$, we can divide the numbers as, $i_1 | | i_2 i_3 | i_4$ Here 0 $i_1 = 1, i_2 = i_3 = 3, i_4 = 4$

Clearly, the number of ways we can separate these numbers, is the number of tuples we need, because numbers with higher indices are always given higher (or equal) values.

So, we can finish with stars-and-bars, as we have $m$ numbers, and $n -1$ separators, so the total number of separations/tuples is: $\binom {m + n - 1}{n - 1}$ which is the same answer as above.

Let $S =\sum_{i_m=1}^n \sum_{i_{m-1}=1}^{i_m} \cdots \sum_{i_2=1}^{i_3} \sum_{i_1=1}^{i_2} 1$

$\displaystyle S=\sum_{i_m=1}^n \sum_{i_{m-1}=1}^{i_m} \cdots \sum_{i_2=1}^{i_3} \binom{i_2}{1}$

$\displaystyle S = \sum_{i_m=1}^n \sum_{i_{m-1}=1}^{i_m} \cdots \sum_{i_3=1}^{i_4} \binom{i_3+1}{2}$

$\displaystyle S = \sum_{i_m=1}^n \sum_{i_{m-1}=1}^{i_m} \cdots \sum_{i_4=1}^{i_5} \binom{i_4+2}{3}$

It's obvious from induction that $\displaystyle S=\sum_{i_m=1}^n \binom{i_m+m-2}{m-1} = \frac{n}{m}\binom{m+n-1}{n}$

Solution to Problem 21:

We use the well known generating function of Central Binomial coefficient:

$\frac{1}{\sqrt{1 - 4x}} = \sum_{n=0}^\infty \frac{(2n)!}{(n!)^2}x^n$

Here substitute $x=\frac{1}{5}$.

Hence, $\large \boxed{\displaystyle \sum_{n=0}^{\infty} \binom{2n}{n} 5^{-n} = \sqrt{5}}$

Define $\displaystyle T(k) = \sum_{m=1}^{n-1} \frac{\cos{\frac{mk\pi}{n}}}{1+\cos{\frac{m\pi}{n}}}$

Observing that $\cos{(k+1)\theta} + 2\cos{k\theta} + \cos{(k-1)\theta} = 2\cos{k\theta}(1+\cos{\theta})$

The following recurrence relation is obtained:

$\displaystyle T(k+1) + 2T(k) + T(k-1) = 2 \sum_{m=1}^{n-1} \cos{\frac{mk\pi}{n}} = 1+\cos{k\pi} + \sin{k\pi} \tan{\frac{k\pi}{2n}}$

Note that an implicit restriction of $k \notin n \mathbb{Z}$ is forced because the GP formula is invalid when the common ratio is $\pm 1$.

Evaluate the recurrence relation at $k+1$ and add together to yield

$T(k+2) + 3T(k+1) + 3T(k) + T(k-1) = 2$

Evaluate this recurrence relation at $k$ and take the difference to homogenise the recurrence relation:

$T(k+2) + 2T(k+1) - 2T(k-1) - T(k-2) = 0$

The characteristic roots are $1$ and $-1$ with multiplicity $3$.

The general solution to the recurrence relation is thus

$(C_1 k^2 + C_2 k+C_3)(-1)^k + C_4$

Returning to the original problem, define $\displaystyle S(a,b,n) = \sum_{m=1}^{n-1} \frac{\sin{\frac{am\pi}{n}}\sin{\frac{bm\pi}{n}}}{1+\cos{\frac{m\pi}{n}}}$ where $0 \le b \le a \le n$

We find that $2S(a,b,n) = T(a-b)-T(a+b)$

Note that $(-1)^{a+b} \equiv (-1)^{a-b}$

Substituting $T(k)$ into the above and expanding, we obtain

$\displaystyle S(a,b,n) = (2C_1 a + C_2)b(-1)^{a+b+1} = (-1)^{a+b}\hat{C_1}b(\hat{C_2}-a)$

A short computation yields $S(1,1,n) = n-1$

From $S(n,b,n) = 0 \, \& \, S(1,1,n)>0$, $\hat{C_2} = n$

From $S(1,1,n) = n-1$, $\hat{C_1} = 1$

Therefore, we conclude $S(a,b,n) = (-1)^{a+b} b(n-a)$

Substituting $b=17, a=39, n=100$, and the result which was to be shown follows.

$\displaystyle \sum_{n\geq -\infty} \frac{(-1)^n}{(n+a)^4}=\frac{(-1)^4}{4^2}\left( \sum_{k \geq 0}\frac{1}{(k+1-\frac{a}2)^4}+\frac{1}{(k+\frac{a}2)^4}-\frac{1}{(k-\frac{1-a}2)^4}-\frac{1}{(k-\frac{1+a}2)^4} \right)$

=$\frac{1}{16} (\frac16(\psi^{(3)}(1-\frac{a}2)+\psi^{(3)}(\frac{a}2)-\psi^{(3)}(\frac{1-a}2)-\psi^{(3)}(\frac{1+a}2)))$

Using reflection formula and On a little bit solving you get

$\frac{\pi^4}{6}\frac{cos(\pi a)}{\sin^2(\pi a)}(6\csc^2(\pi a)-1)$ Sorry for short solution

$\text{Alternative :}$

Consider the function $\displaystyle f(z)=\frac{1}{(z+a)^4}$ which analytic and has poles at the point $z=-a$ of order 4.

Considering the square contour positively oriented having vertices $\pm (1+i)$ and by residue lemma we derive ,

$\displaystyle Res(-a) = \frac{1}{6} \lim_{z\to -a} \frac{d^3}{dz^3}(π\csc(πz))$

Finally using alternate summation theorem ,

$\displaystyle \sum_{n=-\infty}^{\infty} \frac{(-1)^n}{(n+a)^4} =- \frac{1}{6} \lim_{z\to -a} \frac{d^3}{dz^3}(π\csc(πz)) = \frac{\pi^4}{6} \frac{\cos(\pi a)}{\sin^2 (\pi a)} (6\csc^2 (\pi a) -1)$

which gives the result as desired.

Using Summation by parts $\displaystyle \sum _{ k=1 }^{ n }{ { H }_{ k }^{ 2 } } = H_n((n+1)H_n-n)-\sum_{1\leq k \leq n-1}\frac{(k+1)H_k-k}{k+1}$

$=\displaystyle (n+1)H_n^2-nH_n-nH_{n-1}+n-1+n-H_n$

Partial fractions used here: $\frac { { 32n }^{ 3 }-{ 24n }^{ 2 }+8n-1 }{ \left( 2n+1 \right) { \left( n \right) }^{ 4 }{ \left( 2n-1 \right) }^{ 4 } } =-\frac { 1 }{ { n }^{ 4 } } +\frac { 2 }{ { n }^{ 3 } } -\frac { 4 }{ { n }^{ 2 } } -\frac { 1 }{ { 2n }-1 } -\frac { 15 }{ 2{ n }+1 } +\frac { 2 }{ { \left( 2n-1 \right) }^{ 2 } } -\frac { 4 }{ { \left( 2n-1 \right) }^{ 3 } } +\frac { 8 }{ { \left( 2n-1 \right) }^{ 4 } } +\frac { 8 }{ { n } }$

Formula (1) used: ${ H }_{ \frac { q }{ p } }^{ (m) }=\zeta \left( m \right) -{ p }^{ m }\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { \left( q+pn \right) }^{ m } } }$

Formula (2) used: $\sum _{ k=1 }^{ p }{ \frac { 1 }{ 2k+1 } } =\frac { 1 }{ 2 } \left( \Psi \left( p+\frac { 3 }{ 2 } \right) -\Psi \left( \frac { 3 }{ 2 } \right) \right)$