## Problem 1

Find a closed form solution for $f(z) = \sum_{n=1}^\infty \sum_{k=1}^n \frac{1}{k}z^n = z + \frac{3}{2}z^2 + \frac{11}{6}z^3 + \dots$.

### Solution

We have

\begin{align*} f(z) - zf(z) &= z + \frac{z^2}{2} + \frac{z^3}{3} + \dots \\ &= \int \frac{1}{1-z}dz = -ln(1-z) \\ \implies f(z) &= -\frac{ln(1-z)}{1-z} \end{align*}

## Problem 2

Evaluate $\sum_{n=0}^{\infty} \frac{(-1)^n)}{3^n(2n + 1)}$.

### Solution

We have,

\begin{align*} \sum_{n=0}^{\infty} \frac{x^{2n+1}}{2n+1} &= \int \frac{1}{1-x^2} dx \\ &= \frac{1}{2} (log(1+x) - log(1-x)) \end{align*}

Dividing both sides of the above equation by $x$ and substituting $x = \frac{i}{\sqrt{3}}$, we get

\begin{align*} \sum_{n=0}^{\infty} \frac{(-1)^n}{3^n(2n + 1)} &= \frac{\sqrt{3}}{2i} \log \left(\frac{\sqrt{3}+i}{\sqrt{3}-i} \right)\\ &= \frac{\sqrt{3}}{2i} \frac{i2\pi}{6} = \frac{\pi}{2\sqrt{3}} \end{align*}

## Problem 3

Compute

\begin{align*} \sum_{k=1}^{101} \frac{a_k^3}{1-3a_k+3a_k^2} \end{align*}

where $a_k = \frac{k}{101}$, $k=1,2,\dots,101$.

### Solution

Let $f(x) = \frac{x^3}{1-3x+3x^2}$, we have $f(1-x) + f(x) = 1$, that means $f(a_k) + f(a_{101-k})=1$.Therefore the above the above sum is $50 + f(\frac{101}{101})=51$.

## Problem 4

Evaluate $1 + \frac{1^2}{2!} + \frac{2^2}{3!} + \frac{3^2}{4!} + \dots$.

### Solution

We have

$$\frac{e^x - 1}{x} = \sum_{k=1}^\infty \frac{x^{k-1}}{k!}$$

Differentiating both sides, we get

$$\frac{xe^x- (e^x-1)}{x^2} = \sum_{k=2}^\infty \frac{(k-1)x^{k-2}}{k!}$$

Multiplying by $x$ and differentiating both sides, we get

$$e^x - \left(\frac{xe^x- (e^x-1)}{x^2}\right)= \sum_{k=2}^\infty \frac{(k-1)^2x^{k-2}}{k!}$$

Therefore,

$$1 + \frac{1^2}{2!} + \frac{2^2}{3!} + \frac{3^2}{4!} + \dots = \bf{e}.$$