Question

find the sum of infinite series..

Answer 1

1/3 + 1/7 + 1/13 +.....infinity = ?

Answer 2

@jim_thompson5910

Answer 3

@Ashleyisakitty

Answer 4

Does it give a rule how the terms are generated?

Answer 5

1/3 = 1/(1+1+1^2)
1/7 = 1/(1+2+2^2)
1/13 = 1/(1+3+3^2)

Answer 6

so the nth term looks like

1/(1+n+n^2)

Answer 7

yes..

Answer 8

when I type
`sum(1/(1+n+n^2),n=1..infinity)`
into wolfram alpha, I get this
http://www.wolframalpha.com/input/?i=sum(1%2F(1%2Bn%2Bn%5E2),n%3D1..infinity)
and honestly, I've never even heard of the `digama` function til now, so I have no clue. Testing to see if a series converges or not is relatively easy compared to actually finding the infinite sum.

Answer 9

but it turns out that
\[\Large \frac{1}{3}+\frac{1}{7}+\frac{1}{13}+\ldots=\sum_{n=1}^{\infty}\frac{1}{1+n+n^2} \approx 0.798147\]
(according to wolfram alpha)
so that's at least something I guess

Answer 10

the answer given in the solution is 1
actually i learnt to solve such series nearly a year ago but it seems i forgot it.....
but anyways thanks... i will look for it again

Answer 11

hmm then there may be a term missing or a typo somewhere

Answer 12

may be but i think the given series is correct