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I am currently studying for the actuarial exam FM. I have been working on problems involving duration and convexity of perpetuities. I was in the middle of a problem, and got stuck on trying to evaluate this series:

$$\sum_{n=1}^{\infty} n^2(1.05)^{-n}$$

On my upcoming exam, I am only allowed to use a TI-30XS Multiview calculator, which has most functions (log, exponential, trignometric, etc), but does not do summation. Normally, I would plug something like this into my TI-84 calcultor and find the partial sum till 100, 1,000, 10,000, etc and try to see if the series converges. Doing this gives me $17,220$ as the value, which wolframalpha confirms is correct.

I've also had this issue with this series

$$\sum_{n=1}^{\infty} (1.05)^{n}(1.10)^{-n}(n)$$ (which wolframalpha confirms is equal to 462).

Can I evaluate these types of series without using an advanced calculator? If so, how? Note, the solutions given for both of these problems found ways to sidestep these sums, but I am curious about if they can be evaluated directly.

Thanks in advance.

AfronPie
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    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. – Maximilian Janisch Dec 02 '21 at 15:56

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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 $$\sum_{n=1}^{\infty} n^2(1.05)^{-n}=f(\frac{1}{1.05})=\cdots.$$

xpaul
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  • Good one (but very well-known). I'll add a discrete analogue soon. – Maximilian Janisch Dec 02 '21 at 16:03
  • Switching of sum and integral is justified using Fubini, I assume? – Maximilian Janisch Dec 02 '21 at 16:09
  • @xpaul Thanks for the answer. I just went through all the steps you outlined and came up with 17,220! I would have never have thought about taking the derivative of the $nx^n$ series. Does this method require having $\sum_{n=1}^\infty nx^n=\frac{x}{(1-x)^2}$ memorized? – AfronPie Dec 02 '21 at 17:01
  • You don't need to memorize. Take derivative for $\frac1{1-x}=\sum_{n=0}^\infty x^n$ and then you will get it. – xpaul Dec 02 '21 at 17:36
  • @MaximilianJanisch, you can use M-test. – xpaul Dec 02 '21 at 17:46
  • @xpaul Oh, I see now. I like this method:) – AfronPie Dec 02 '21 at 18:17
  • @xpaul I am not sure if I follow: Weierstrass M-test guarantees convergence but not that two limits can be exchanged. – Maximilian Janisch Dec 02 '21 at 19:16
  • @MaximilianJanisch Use M-test in $|x|\le r<1$ for uniformly convergence of $\sum f_n(x)$ and $\sum f'_n(x)$. See https://www.math.ucla.edu/~tao/resource/general/131bh.1.03s/week45.pdf – xpaul Dec 03 '21 at 00:28
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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.

Maximilian Janisch
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  • Thanks for the answer @Maximilian. I have actually never heard of summation by parts until reading about it in your answer. I follow what you are saying but at the end when you substitute into the summation by parts formula you imply that $A_{N}b_{N}=\frac{\rho(\rho^N-1)}{\rho-1}$, but isn't $A_{N}b_{N}=p^{n}n^{2}$ (the nth term)? – AfronPie Dec 02 '21 at 17:12