Summation 2

We note that the Maclaurin series of

$\begin{aligned} \ln (1+x) & = x - \frac {x^2}2 + \frac {x^3}3 - \cdots = \sum_{n=1}^\infty \frac {(-1)^{n+1}x^n}n \\ \sum_{n=1}^\infty \frac {(-1)^n}n & = - \sum_{n=1}^\infty \frac {(-1)^{n+1}x^n}n \bigg|_{x=1} = - \ln (1+1) = - \ln 2 \\ \implies P & = - \ln 2 \end{aligned}$

Using Maclaurin series again,

$\begin{aligned} \sum_{n=0}^\infty \frac {x^n}{n!} & = e^x& \small \blue{\text{Differentiate both sides w.r.t. }x} \\ \sum_{n=0}^\infty \frac {nx^{n-1}}{n!} & = e^x & \small \blue{\text{Multiply both sides by }x} \\ \sum_{n=0}^\infty \frac {nx^n}{n!} & = xe^x & \small \blue{\text{Differentiate both sides w.r.t. }x} \\ \sum_{n=0}^\infty \frac {n^2x^{n-1}}{n!} & = e^x + xe^x & \small \blue{\text{Multiply both sides by }x} \\ \sum_{n=0}^\infty \frac {n^2x^n}{n!} & = xe^x + x^2e^x & \small \blue{\text{Differentiate both sides w.r.t. }x} \\ \sum_{n=0}^\infty \frac {n^3x^{n-1}}{n!} & = e^x + xe^x + 2xe^x + x^2e^x & \small \blue{\text{Multiply both sides by }x} \\ \sum_{n=0}^\infty \frac {n^3x^n}{n!} & = xe^x + 3x^2e^x + x^3e^x & \small \blue{\text{Putting }x=1} \\ \sum_{n=0}^\infty \frac {n^3}{n!} & = 5e \\ \implies Q & = 5e \end{aligned}$

Therefore, $\left(\dfrac Q5 \right)^{5P} = e^{-5\ln 2} = \dfrac 1{32} = \boxed{0.03125}$ .

The first one ( $P$ ), just google it up under "alternating sum of reciprocals".

For the second one ( $Q$ ), in language of exponential generating function, defining $Q(x)=\sum_{n=0}^{\infty} \frac{n^3}{n!}x^n$ , you need to calculate $Q(1)$ . Knowing $e^x=\sum_{n=0}^{\infty} \frac{1}{n!}x^n$ , $Q(x)=D^3 \cdot e^x$ , where operator $D$ , that acts on a function of $x$ , is basically taking derivative and multiplying the result by $x$ . There rest is easy!

We are going to use some Taylor expansions here, but before let's take a quick recap.

$\large F(x) = \sum_{n=0}^\infty\frac{[F^{(n)}(a)](x-a)^n}{n!}$

Where $\large x$ is variable, $\large a$ is constant and $F^{(n)}(x-a)$ means the $\large n^{th}$ derivative of the Function.

Now keeping $F(x) = e^x$ [ e is the Euler's number ] and a = 0 , we get the Taylor expansion as :

$\large e^x = \frac{x^0}{0!} + \frac{x^1}{1!} + \frac{x^2}{2!} + ...$

Thus , $\large e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$

From this we can say that $\sum_{n=0}^\infty \frac{1^n}{n!} = e^1 = e$

• Next for the sum

$\large \sum_{n=0}^\infty \frac{n}{n!}$

$\large = \frac{0}{0!} + \sum_{n=1}^\infty \frac{n}{n!} = \sum_{n=1}^\infty \frac{1}{(n-1)!}$

$\large = \sum_{n=0}^\infty \frac{1}{n!} = e$

• Similarly for

$\large \sum_{n=0}^\infty \frac{n²}{n!}$

$\large = \frac{0}{0!} + \sum_{n=1}^\infty \frac{n²}{n!} = \sum_{n=1}^\infty \frac{n}{(n-1)!}$

$\large = \sum_{n=0}^\infty \frac{n+1}{n!}$

$\large = \sum_{n=0}^\infty \frac{n}{n!} + \sum_{n=0}^\infty \frac{1}{n!} = e + e$

$\large= 2e$

• At last for $\large Q$

$\large \sum_{n=0}^\infty \frac{n³}{n!}$

$\large = \frac{0}{0!} + \sum_{n=1}^\infty \frac{n³}{n!} = \sum_{n=1}^\infty \frac{n²}{(n-1)!}$

$\large = \sum_{n=0}^\infty \frac{(n+1)²}{n!}$

$\large = \sum_{n=0}^\infty \frac{n²+2n+1}{n!}$

$\large = (2e) + 2(e) + 1(e) = 5e$

Thus $\large Q= 5e$

Next if we take $F(x) = Ln(1+x)$ [ Ln is the Natural Logarithm ] , $-1 ≤ x ≤ 1$ and a = 0 , we get the Taylor expansion as :

$\large Ln(1+x) = \frac{x}{1} - \frac{x²}{2} + \frac{x³}{3} - \frac{x⁴}{4} + ...$

So , $Ln(1+x) = \sum_{n=1}^\infty \frac{-(x^n)(-1)^n}{n}$

If we keep $x = 1$ here , then we get the following result :

$Ln(1+1) = \sum_{n=1}^\infty \frac{-(1^n)(-1)^n}{n}$

$Ln(2) = \sum_{n=1}^\infty \frac{-(-1)^n}{n}$

Or we can say $\large P = -Ln(2)$

Now we need to find the value of $\large (Q/5)^{(5P)}$

$\large = ((5e)/5)^{(5(-Ln 2))}$

$\large = (e)^{(Ln (2^{-5}))}$

$\large = 2^{-5} = \frac{1}{32}$

$\large = 0.03125$