Have You Heard Of Euler's Sum?

$\underline{\text{Method 1}}$

Note that,

$\displaystyle \int_{0}^{1} x^{n-1} \mathrm{d}x = \dfrac{1}{n}$

Differentiating w.r.t. to $n$ , $(p-1)$ times, we get,

$\displaystyle \dfrac{1}{n^{p}} = \dfrac{(-1)^{p-1}}{(p-1)!} \int_{0}^{1} x^{n-1} [\ln(x)]^{p-1} \mathrm{d}x$

$\displaystyle \implies \sum_{n=1}^{\infty} \dfrac{H_{n}}{n^{p}} = \dfrac{(-1)^{p-1}}{(p-1)!} \int_{0}^{1} [\ln(x)]^{p-1} \sum_{n=1}^{\infty} H_{n} x^{n-1} \mathrm{d}x$

Since $\displaystyle \sum_{n=1}^{\infty} H_{n} x^{n} = -\dfrac{\ln(1-x)}{1-x}$ , we get,

$\displaystyle \text{S} = \dfrac{(-1)^{p}}{(p-1)!} \int_{0}^{1}\dfrac{[\ln(x)]^{p-1} \cdot \ln(1-x) }{x(1-x)} \mathrm{d}x$

Recall the Beta Function $\displaystyle \operatorname{B}(a,b) = \int_{0}^{1} x^{a-1} (1-x)^{b-1} \mathrm{d}x = \dfrac{\Gamma(a) \Gamma(b)}{\Gamma(a+b)}$

$\displaystyle \implies \text{S} = \dfrac{(-1)^{p}}{(p-1)!} \lim_{a \to 0^+} \lim_{b \to 0^+} \dfrac{{\partial}^{p-1}}{\partial a^{p-1}} \left( \dfrac{\partial}{\partial b} \operatorname{B}(a,b) \right)$

$\displaystyle = \left(1+\dfrac{p}{2} \right)\zeta(p+1)-\dfrac{1}{2}\sum_{k=1}^{p-2}\zeta(k+1)\zeta(p-k)$

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$\underline{\text{Method 2}}$

Proposition : $\displaystyle \int_{0}^{1} x^{n} \operatorname{Li}_{s}(x) \mathrm{d}x = (-1)^{s+1} \dfrac{H_{n+1}}{(n+1)^{s}} + \sum_{r=2}^{s} \left[(-1)^{s+r} \dfrac{\zeta(r)}{(n+1)^{s-r+1}} \right] \ ; \ s \geq 2 \ ; \ s,n \in \mathbb{Z^{+}}$

Proof : I have proved it here

Now,

$\displaystyle \sum_{n=1}^{\infty} \dfrac{H_{n}}{n^{p}} = \sum_{n=1}^{\infty} \int_{0}^{1} \dfrac{1}{n^p} \left(\dfrac{1-x^n}{1-x}\right) \mathrm{d}x$

$\displaystyle = \int_{0}^{1} \dfrac{\zeta(p) - \operatorname{Li}_{p}(x)}{1-x} \mathrm{d}x$

$\displaystyle \sum_{r=1}^{\infty} \int_{0}^{1} (x^r \zeta(p) - x^r \operatorname{Li}_{p}(x)) \mathrm{d}x$

Let $\displaystyle \text{S} = \sum_{n=1}^{\infty} \dfrac{H_{n}}{n^{p}}$

Using the proposition,

$\displaystyle \text{S} = \sum_{r=1}^{\infty} \left[ \dfrac{\zeta(p)}{r} - (-1)^{p+1} \dfrac{H_{r+1}}{(r+1)^{p}} + \sum_{k=2}^{p} \left((-1)^{p+k} \dfrac{\zeta(k)}{(n+1)^{p-k+1}}\right) \right]$

$\displaystyle = (-1)^p \sum_{r=1}^{\infty}\dfrac{H_{r}}{r^p} + (-1)^{p}\sum_{r=0}^{\infty} \sum_{k=2}^{p-1} \dfrac{(-1)^{k} \zeta(k)}{(r+1)^{p-k+1}}$

If $p$ is odd, we get,

$\displaystyle 2\text{S} = \sum_{k=1}^{p-2} (-1)^{k-1} \zeta(k+1) \zeta(p-k)$

$\displaystyle \implies \text{S} = \dfrac{1}{2} \sum_{k=1}^{p-2} (-1)^{k-1} \zeta(k+1) \zeta(p-k)$

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$\underline{\text{Method 3}}$

Let $p$ be odd, $p \geq 3$ ,

$\displaystyle \text{S} = \sum_{n=1}^{\infty} \dfrac{H_{n}}{n^{p}}$

$\displaystyle = \int_{0}^{1} \dfrac{\zeta(p) - \operatorname{Li}_{p}(x)}{1-x} \mathrm{d}x$

Using I.B.P., we have,

$\displaystyle \text{S} = - \left[\Big(\zeta(p) - \operatorname{Li}_{p}(x) \Big) \ln(1-x)\right]_{0}^{1} - \int_{0}^{1} \dfrac{\log(1-x) \operatorname{Li}_{p-1}(x)}{x} \ \mathrm{d}x$

$\displaystyle = - \int_{0}^{1} \dfrac{\log(1-x) \operatorname{Li}_{p-1}(x)}{x} \mathrm{d}x$

Again, using I.B.P., we have,

$\displaystyle \text{S} = \left[\operatorname{Li}_{2}(x) \text{Li}_{p-1}(x) \right]_{0}^{1} - \int_{0}^{1} \dfrac{\operatorname{Li}_{2}(x) \operatorname{Li}_{p-2}(x)}{x} \mathrm{d}x$

$\displaystyle = \zeta(2) \zeta(q-1) - \int_{0}^{1} \dfrac{\operatorname{Li}_{2}(x) \text{Li}_{p-2}(x)}{x} \ \mathrm{d}x$

Similarly,applying I.B.P. $p$ times, we get,

$\displaystyle \implies \text{S} = \sum_{k=1}^{p-2} (-1)^{k-1} \zeta(k+1) \zeta(p-k) + \int_{0}^{1} \dfrac{\log(1-x) \operatorname{Li}_{p-1}(x)}{x} \ \mathrm{d}x$

$\displaystyle = \sum_{k=1}^{p-2} (-1)^{k-1} \zeta(k+1) \zeta(p-k) - \int_{0}^{1} \dfrac{\zeta(p) - \operatorname{Li}_{p}(x)}{1-x} \mathrm{d}x$

$\displaystyle \implies \text{S} = \dfrac{1}{2}\sum_{k=1}^{p-2} (-1)^{k-1} \zeta(k+1) \zeta(p-k)$

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Thus,

$\displaystyle \sum_{n=1}^{\infty} \dfrac{H_{n}}{n^5} = \dfrac{\pi^{6}}{540} - \dfrac{\zeta^2(3)}{2}$

$\therefore \text{Ans.} = \boxed{553}$

This is an interesting evaluation, For this let us consider some sums,

$\text{Lemma 1 :}$

$\displaystyle \alpha(p) = \sum_{n,m=1}^{\infty} \frac{1}{n^{p-1}m(n-m)} = \sum_{k=1}^{\infty} \frac{H_k}{k^p} - 2\zeta(p+1)$

$\text{Proof :}$

$\displaystyle \sum_{n,m=1}^{\infty} \frac{1}{n^{p-1}m(n-m)} = \sum_{n,m=1}^{\infty} \frac{1}{n^p}\frac{n}{m(n-m)}$

$\displaystyle \alpha(p) = \sum_{n=1}^{\infty} \frac{1}{n^p} \sum_{m=1}^{\infty}(\frac{1}{m}+\frac{1}{n-m})$

$\displaystyle \alpha(p) = \sum_{n=1}^{\infty} \frac{1}{n^p}(H_n-\frac{2}{n})$

$\displaystyle \alpha(p)= \color{#D61F06}{\sum_{k=1}^{\infty} \frac{H_k}{k^p}-2\zeta(p+1)}$

$\text{Lemma 2:}$

Let $m$ be an arbitrary positive integer such that ,

$\displaystyle \beta(p,x) = \sum_{n=1\ne m}^{\infty} \frac{1}{n^p(n-x)} = \frac{1}{x^{p-1}}\sum_{n\ge 1} \frac{1}{n(n-x)} - \sum_{k=1}^{p-1} \frac{\zeta(p-k+1) - \frac{1}{m^{p-k+1}}}{x^k}$

$\text{Proof :}$

$\displaystyle \beta(p,x)= -\frac{1}{x}\sum_{n=1}^{\infty} \frac{n-x-n}{n^p(n-x)} = \frac{1}{x}(\beta(p-1,x)-\sum_{n=1}^{\infty} \frac{1}{n^p})$

We have $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^p} = \zeta(p)-\frac{1}{m^p}$ so the above equation turns,

$\displaystyle \beta(p,x) = \frac{1}{x}(\beta(p-1,x)-\zeta(p)+\frac{1}{m^p})$

With this procedure repeated $(p-2)$ times we get our result,

$\displaystyle \beta(p,x) = \color{#3D99F6}{\sum_{n=1\ne m}^{\infty} \frac{1}{n^p(n-x)} = \frac{1}{x^{p-1}}\sum_{n\ge 1} \frac{1}{n(n-x)} - \sum_{k=1}^{p-1} \frac{\zeta(p-k+1) - \frac{1}{m^{p-k+1}}}{x^k}}$

Now Set $p=p-1$ & $x=m$ we get,

$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{p-1}(n-m)} = \frac{1}{m^{p-1}}\sum_{n=1}^{\infty}\frac{m}{n-m} - \sum_{k=1}^{p-2} \frac{\zeta(p-k+1) - \frac{1}{m^{p-k+1}}}{m^k}$

Dividing both sides by $m$ we get ,

$\displaystyle \alpha(p)= -\sum_{k=1}^{p-2}(\zeta(k+1)\zeta(p-k) - \zeta(p+1)) - \alpha(p)$

Solving and substituting for $\displaystyle \alpha(p)$ we derive ,

$\displaystyle \color{#20A900}{ \sum_{n=1}^{\infty} \frac{H_n}{n^p} = \frac{1}{2}(p+2)\zeta(p+1)-\frac{1}{2}\sum_{k=1}^{p-2}\zeta(k+1)\zeta(p-k)}$

Hence We are to evaluate $\displaystyle \alpha(5) = \sum_{n=1}^{\infty}\frac{H_n}{n^5} = \frac{\pi^6}{540}-\frac{\zeta^2(3)}{2}$ , which makes the answer $\boxed{540+6+2+3+2=553}$