Sum of the following series are:?

Question

1.) 1^2 - (2^2/5) + (3^2/5^2) - (4^2/5^3) +(5^2/5^4) - (6^2/5^5) ...... till infinity
2.) 1/(1+1^2+1^4) + 2/(1^2+2^2+2^4) + 3(1+3^2+3^4) ........ till infinity

Please show the steps or else I will not understand how you arrived at the answer.
Can you also give tips on how to solve these types of AP, GP, HP questions

kb · Accepted Answer

1) This equals Σ(n = 1 to ∞) (-1)^(n-1) n^2 / 5^(n-1)
= Σ(n = 1 to ∞) n^2 (-1/5)^(n-1)

To sum this, start with the geometric series
1/(1 - x) = Σ(n = 0 to ∞) x^n, valid for |x| < 1.

Differentiate both sides:
1/(1 - x)^2 = Σ(n = 1 to ∞) nx^(n-1), valid for |x| < 1.

Multiply both sides by x:
x/(1 - x)^2 = Σ(n = 1 to ∞) nx^n; note the extra factor of n.

So, differentiate both sides again:
(1 + x)/(1 - x)^3 = Σ(n = 1 to ∞) n^2 x^(n-1), valid for |x| < 1.

Let x = -1/5 (which does satisfy |x| < 1):
Σ(n = 1 to ∞) n^2 (-1/5)^(n-1) = (4/5)/(6/5)^3 = 25/54.

Double check:
http://www.wolframalpha.com/input/?i=%CE%A3%28n+%3D+1+to+%E2%88%9E%29+%28-1%29^%28n-1%29+n^2+%2F+5^%28n-1%29
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2) Rewrite as a telescoping sum.

Σ(n = 1 to ∞) n/(1 + n^2 + n^4)
= Σ(n = 1 to ∞) n/((n^4 + 2n^2 + 1) - n^2)
= Σ(n = 1 to ∞) n/((n^2 + 1)^2 - n^2)
= Σ(n = 1 to ∞) n/[(n^2 + n + 1)(n^2 - n + 1)]
= Σ(n = 1 to ∞) (1/2) [1/(n^2 - n + 1) - 1/(n^2 + n + 1)], by partial fractions

Expanding this:
(1/2) * lim(k→∞) Σ(n = 1 to k) [1/(n^2 - n + 1) - 1/(n^2 + n + 1)]
= (1/2) * lim(k→∞) [(1 - 1/3) + (1/3 - 1/7) + (1/7 - 1/13) + ... + (1/(k^2 - k + 1) - 1/(k^2 + k + 1))]
= (1/2) * lim(k→∞) [(1 - 1/3) + (1/3 - 1/7) + (1/7 - 1/13) + ... 
+ (1/(k^2 - k + 1) - 1/((k+1)^2 - (k+1) + 1))]

= (1/2) * lim(k→∞) (1 - 1/((k+1)^2 - (k+1) + 1)), since all terms cancel in pairs
= (1/2) * (1 - 0)
= 1/2.

I hope this helps!

Sum of the following series are:?

1 Answer