Solve Infinite Summations w/o Pi: Tips & Tricks

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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:
\sum_{n=1}^{\infty}\frac{n+1}{6^n}

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,
\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}.

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):
\sum_{n=1}^{\infty} n x^{n-1} = \frac{1}{(1-x)^2}.

Can you take it from here?
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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