# Determine if this infinite sum ( 2^(2n+1)/[3^(2n+1) * (2n+1)] converges or diverges.?

Determine if (sum from n = 2 to infinity) 2^(2n+1)/[3^(2n+1) * (2n+1)] converges or diverges. If it converges, find the value that it converges to. If it diverges, justify your conclusion.

Any help would be much appreciated!!

I tried using the ratio test to prove convergence, but that doesn't help me find the value that it converges to...

### 1 Answer

- kbLv 79 years agoBest answer
Note that we can rewrite this sum as

Σ(n=2 to ∞) (2/3)^(2n+1) / (2n+1).

So, it suffices to compute (by letting x = 2/3)

Σ(n=2 to ∞) x^(2n+1) / (2n+1).

-----------------------------

Start with the geometric series

1/(1 - x) = Σ(n=0 to ∞) x^n, convergent for |x| < 1.

Replace x with x^2:

1/(1 - x^2) = Σ(n=0 to ∞) x^(2n), convergent for |x^2| < 1 ==> |x| < 1.

Since 1/(1 - x^2) = (1/2) [-1/(x - 1) + 1/(x + 1)], integrating both sides from 0 to x yields

(1/2) [ln|x + 1| - ln|x - 1|] = Σ(n=0 to ∞) x^(2n+1)/(2n+1), still convergent for |x| < 1.

Hence,

Σ(n=0 to ∞) x^(2n+1)/(2n+1) = (1/2) ln |(x + 1) / (x - 1)|

Therefore, letting x = 2/3 (noting that |2/3| < 1) yields

Σ(n=0 to ∞) (2/3)^(2n+1)/(2n+1) = (1/2) ln |(2/3 + 1) / (2/3 - 1)| = (1/2) ln 5.

==> Σ(n=2 to ∞) (2/3)^(2n+1) / (2n+1)

= (1/2) ln 5 - [2/3 + (2/3)^3/3]

= (1/2) ln 5 - 62/81.

I hope this helps!