Find the sum of the series

Question

 ∑1 to ∞

(4/n(n+1)+5/2^n) 

Answer 1

I'll have to assume you mean 4/(n(n+1)). In the future, please be mindful of order of operations to avoid ambiguity.

∑ [n=1,∞] 4/(n(n+1)) + 5/2^n

(∑ [n=1,∞] 4/(n(n+1)) ) + (∑ [n=1,∞] 5/2^n)

(∑ [n=1,∞] 5/2^n ) = 5*(1/2+1/4+1/8+...) = 5*1 = 5

The sum 1/2+1/4+1/8+... is well known to be 1. A simple proof is that, if you double it, you get 1+1/2+1/4+..., which is the same as the initial sum + 1. If double the sum is the same as the sum + 1, then the sum is 1.

∑ [n=1,∞] 4/(n(n+1)) = ∑ [n=1,∞] 4((n+1)-n) / (n(n+1)) ) = ∑[n=1,∞] 4/n - 4/(n+1)

Let's take the limit as the upper limit of n goes to ∞:

lim W→∞ ∑ [n=1,W] 4/n - 4/(n+1)

lim W→∞ (∑ [n=1,W] 4/n) - (∑ [n=1,W] 4/(n+1))

lim W→∞ (∑ [n=1,W] 4/n) - (∑ [n=2,W+1] 4/n)

lim W→∞ 4/1 + (∑ [n=2,W] 4/n) - (∑ [n=2,W] 4/n) - 4/(W+1)

lim W→∞ 4 - 4/(W+1)

4

Therefore, our total sum is just 4 + 5 = 9

Answer 2

4 / [n(n+1)] = A/n + B/(n+1)

4 = A(n+1) + Bn

4 = A + (A+B)n

4 + 0n = A + (A+B)n

A = 4 and A+B = 0

So, A = 4 and B = -4

∑_{(n=1 to ∞)} (4/[n(n+1)) = 4∑_{(n=1 to ∞)} [1/n - 1/(n+1)]

S_n = sum of first n terms = 4[(1-1/2) + (1/2-1/3) + (1/3-1/4) + ...+ (1/(n-1)-1/n) + (1/n-1/(n+1)] = 4[1 - 1/(n+1)]

Sum of series = lim_n→∞(S_n) = 4lim_n→∞[1 - 1/(n+1)] = 4

∑_{n=1 to ∞}[5/2ⁿ] = 5∑_{n=1 to ∞}(1/2)ⁿ This is 5 times a geometric series with r = 1/2 and first term 1/2.

= 5[ (1/2) / (1 - 1/2)] = 5

So, ∑_{n=1 to∞} [4/(n(n+1)) + 5/2ⁿ] = ∑_{n=1 to ∞}[4/(n(n+1))] + ∑_{n=1 to ∞}[5/2ⁿ] = 4 + 5 = 9