Evaluating series (such as $\sum_{n=1}^{\infty} n^2(1.05)^{-n}$) [duplicate]

Solution 1:

Hint: Let $$ f(x)=\sum_{n=1}^\infty n^2x^n, |x|<1 $$ and then $$ \int_0^x\frac{f(t)}{t}dt=\sum_{n=1}^\infty nx^n=\frac{x}{(1-x)^2}. $$ So $$ f(x)=x\bigg(\frac{x}{(1-x)^2}\bigg)'=\cdots. $$ So $$\sum_{n=1}^{\infty} n^2(1.05)^{-n}=f(\frac{1}{1.05})=\cdots.$$

Solution 2:

You are looking at a sum of the form $$\sum_{n=1}^\infty n^2 \varrho^n,$$ where, in your case $\varrho=\frac1{1.05}$. Such a series always converges when $\varrho\in]-1,1[$, which is true in your case. It can be evaluated explicitly using summation by parts twice.

Lemma (Summation by parts). Let $N\in\mathbb N$ and let $a_1,\dots, a_N,b_1,\dots,b_N\in\mathbb R$. Then $$\sum_{k=1}^{N}a_k b_k = A_N b_N + \sum_{k=1}^{N-1}A_k(b_k-b_{k+1})$$ where $$A_k \overset{\text{Def.}}=a_1 + a_2 + \ldots + a_k.$$

Proof. Left as an exercise. (Hint: Use induction over $N$.)

In our case: Let $N\in\mathbb N$. Let $a_n=\rho^n$ and $b_n=n^2$. We compute for $\rho\neq 1$ $$A_k = \sum_{n=1}^k \rho^n = \frac{\rho(\rho^k-1)}{\rho-1}.$$ Therefore, by the above Lemma, $$\sum_{n=1}^N \rho^n n^2 = \frac{\rho(\rho^N-1)}{\rho-1} N^2+\sum_{n=1}^{N-1} \frac{\rho(\rho^n-1)}{\rho-1}(-2n-1).$$

Now use summation by parts again for the $n\rho^n$ part and you will get a geometric series. So you can explicitly calculate all the partial sums (and their limit) with this approach.