Show whether the series converges or diverges and find its sum

Question

Show whether the series $$\sum^\infty_{k=1} {k^4 \over 2^k}$$ converges or diverges. In case of convergence, try to determine (and prove) the sum of the series.

My attempt

Using the ratio test:

\begin{align} {|x_n+1| \over|x_n|} &= {\dfrac {(n+1)^4} {2^{n+1}} \over \dfrac {n^4} {2^n}} \\ &= {(n+1)^4 \over 2^n 2^1} {2^n \over n^4} \\ &= {n^4 +4n^3 + 6n^2 +4n+1 \over n^4 \color{red}{\times 2}} \end{align}

We first divide both the numerator and the denominator of $x_n$ by the highest power of $n$, i.e., $n^4$, to get:

$${1 + \dfrac 4 n +\dfrac 6 {n^2} +\dfrac 4 {n^3} +\dfrac 1 {n^4} \over 2}$$

Limit of the numerator: \begin{align} \lim \left( 1 + {4 \over n} + {6 \over n^2} +{4 \over n^3} + {1 \over n^4} \right) &= \lim(1) +4\lim \left( {1 \over n} \right) +6\lim \left( {1 \over n^2} \right) +4\lim \left( {1 \over n^3} \right) +\lim \left( {1 \over n^4} \right) \\ &= 1 +4\times0+ 6\times0+ 4\times0 +0 \\ &= 1 \end{align}

Limit of the denominator: $\lim(2)=2$.

As both numerator and denominator have limits, we can conclude:

\begin{align} \lim \left( {1 +\dfrac 4 n +\dfrac 6 {n^2} +\dfrac 4 {n^3} +\dfrac 1 {n^4} \over 2} \right) &= { \lim \left( 1 +\dfrac 4 n +\dfrac 6 {n^2} +\dfrac 4 {n^3} +\dfrac 1 {n^4} \right) \over \lim(2)} \\ &= {1 \over 2} \\ &=: g <1 \end{align}

As we have $0<|g|<1$, so the series converge and we can determine the sum of series:

\begin{align} S &= {a \over 1- r} \\ &= {{1 \over 2} \over 1 - {1 \over 2}} \\ &= {1 \over 2}\times{2 \over 1} \\ &= 1 \end{align}

Is it correct?

Answer 1

To compute the sum, start with

$$\sum_{k=1}^{\infty}x^k=\frac{x}{1-x}$$ which is valid for $|x|<1$. Differentiate once, to get: $$\sum_{k=1}^{\infty}kx^{k-1}=\frac{1}{(1-x)^2}$$ rewrite this as $$\sum_{k=1}^{\infty}kx^{k}=\frac{x}{(1-x)^2}$$ Repeat the process until you arrive at the desired series, and then put $x=1/2$. The result is $150$.