Solve Infinite Summations w/o Pi: Tips & Tricks

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SUMMARY

This discussion focuses on solving infinite summations that do not yield results involving pi, specifically the series \(\sum_{n=1}^{\infty}\frac{n+1}{6^n}\), which converges to 11/25. The user attempted to compute the sum using a TI-nSpire calculator, but faced limitations due to the absence of a Computer Algebra System (CAS). The conversation highlights the utility of recognizing series as special cases of more general forms, particularly the geometric series, and suggests differentiating the geometric series to derive results for related summations.

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Looking for ways to solve infinite summations, I found an ancient topic here talking about solving infinite summations that come out to answers with pi.
How would I solve an infinite summation that does not come out to an answer with pi?

Such as:
[tex]\sum_{n=1}^{\infty}\frac{n+1}{6^n}[/tex]

The solution is 11/25, btw.
My attempt: I am not really experienced with this area of math, so what I did was put it into my TI-nSpire but it couldn't do it because it's not the CAS version. I plugged in a large number such as 999 terms instead of infinity terms and it came out to the right answer, but I am looking for a more "correct" way to solve the problem. I also thought of finding the sum of a geometric sequence but I realized that doesn't really work for most summations.

Oh, I'm also new to these forums, so hi to everybody! =D

EDIT: The n under the summation should say n=1
 
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Many of these problems can be solved by recognizing that your series is actually a special case of a more general one.

One of the more useful series is the geometric series: If |x|<1,
[tex]\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}.[/tex]

It's not immediately obvious how we can use this to evaluate your series. However, there's a nice trick: Differentiate both sides of the above equation term by term to get (for |x| < 1):
[tex]\sum_{n=1}^{\infty} n x^{n-1} = \frac{1}{(1-x)^2}.[/tex]

Can you take it from here?
 
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