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Question: What is the value of $$\frac{1}{3^2+1}+\frac{1}{4^2+2}+\frac{1}{5^2+3} ...$$ up to infinite terms?

Answer: $\frac{13}{36}$

My Approach: I first find out the general term ($T_n$)$${T_n}=\frac{1}{(n+2)^2+n}=\frac{1}{n^2+5n+4}=\frac{1}{(n+4)(n+1)}=\frac{1}{3}\left(\frac{1}{n+1}-\frac{1}{n+4}\right)$$

Using this, I get,$$T_1=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}\right)$$ $$T_2=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{6}\right)$$ $$T_3=\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}\right)$$

I notice right away that the series does not condense into a telescopic series. How do I proceed further?

General Kenobi
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    it does not? go a few more terms ahead then, for i think it does. – Sarvesh Ravichandran Iyer Jul 01 '20 at 16:13
  • @астонвіллаолофмэллбэрг Indeed$$T_4=\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}\right)$$, which crosses out the $\frac{1}{15}$ obtained in $T_1$, and a similar pattern is observed when I go ahead, but consecutive values do not get crossed out, hence my difficulty in finding the sum of this series up to infinite terms. – General Kenobi Jul 01 '20 at 16:17
  • Literally if you remove parentheses and consider the two series for each of the two terms, they telescope after you have considered the first 3 terms of the first series – Peanut Jul 01 '20 at 16:19
  • I'm really sorry, but I don't understand your answer. I'm sure it is correct, but I can't figure it out! Again, really sorry. – General Kenobi Jul 01 '20 at 16:22
  • @GeneralKenobi I hope the answer below was helpful. +1 – Sarvesh Ravichandran Iyer Jul 01 '20 at 16:22
  • @астонвіллаолофмэллбэрг it really was! – General Kenobi Jul 01 '20 at 16:23

2 Answers2

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You have solved it already. Recognise, that you can factor out $\frac 13$ from all terms. You have an infinite terms of +/- fractions, from which the first positive three ($\frac 12$, $\frac 13$, $\frac 14$) remains in the sum, every other is cancelled out by a negative counterpart with a 3 step gap. Therefore the end result is $\frac 13 (\frac 12 + \frac 13 + \frac 14)$.

Dávid Laczkó
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You were so close of solving it!

The final step is of the form

$$\frac{1}{3}\sum_{n=1}^\infty{\frac{1}{n+1}-\frac{1}{n+4}}$$

Writing down the series

$$\require{cancel} \frac{1}{3}\left[\left(\frac{1}{2}-\cancel{\frac{1}{5}}\right)+\left(\frac{1}{3}-\frac{1}{6}\right)+\left(\frac{1}{4}-\frac{1}{7})\right)+\left(\cancel{\frac{1}{5}}-\frac{1}{8})\right)\right]$$

So we get the first, second and third positive term and the last three, since we are evaluating it up to infinity, these last ones will be zero and we end up with

$$\frac{1}{3}\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)=\frac{1}{3}\left(\frac{13}{12}\right)=\boxed{\frac{13}{36}}$$

Samuel A. Morales
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