Solve Infinite Summations w/o Pi: Tips & Tricks

  • Thread starter Thread starter mathg33k
  • Start date Start date
  • Tags Tags
    Infinite
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 13K views
mathg33k
Messages
6
Reaction score
0
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
 
Last edited:
Physics news on Phys.org
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?
 
  • Like
Likes   Reactions: 1 person