Question

Use Theorem 9.4.3 to find the sum of the series.
∞
Σ ( (1/ 5^k) - (1/ (k (k+1) ) )
k=1

Theorem 9.4.3 (slide 24/ 72): slideplayer.com/slide/8024349/

Answer 1

This theorem just says the summation is a linear operation. As are addition and subtraction. So, your sum is just

Σ 1/5^k - Σ 1/(k(k+1))
Σ 1/5^k is just a geometric series.
1/(k(k+1)) = 1/k - 1/(k+1) so the sum is 1 - 1/2 + 1/2 - 1/3 + ... = 1

Answer 2

I have no idea what "Theorem 9.4.3" is in your text book
but let's take it in parts
∞
Σ 1/ 5^k
k = 1
= 1/5 + 1/5^2 + 1/5^3 + ...
this is a geometric series with a = 1/5 and r = 1/5
sum∞ = a/(1-r) = (1/5)/(4/5) = 1/4

the second part is
Σ 1/ (k (k+1) ) , for k = 1 to ∞

By partial fractions i can replace 1/ (k (k+1) )
with 1/k - 1/(k+1)
so Σ 1/ (k (k+1) ) , for k = 1 to ∞
= Σ 1/k - 1/(k+1) ) , for k = 1 to n
= 1/1-1/2 + 1/2-1/3 + 1/2-1/4 + .... + 1/n - 1/(n+1_
= 1 - 1/(n+1) , notice all those other terms cancel

so when n ----> ∞ , 1/(n+1) ----> 0

thus:
Σ 1/ (k (k+1) ) , for k = 1 to ∞
= 1

and finally
∞
Σ ( (1/ 5^k) - (1/ (k (k+1) ) )
k=1
= 1/4 - 1
= -3/4