Verified answer

Split this up into series as follows:

sum(n=1 to infinity) {(1/3)^n - 1/[(n + 1)(n + 2)]}

= sum(n=1 to infinity) (1/3)^n - sum(n=1 to infinity) 1/[(n + 1)(n + 2)].

The first series is geometric and the second is a telescoping series (more on that later).

The first series is an infinite geometric series with a first term of 1/3 and a common ratio of 1/3. Since the sum of an infinite geometric series with a first term of a and a common ratio of r is a/(1 - r), we see that:

sum(n=1 to infinity) (1/3)^n = (1/3)/(1 - 1/3) = (1/3)/(2/3) = 1/2.

To find the sum of the second series, consider the sum of the first k terms.

sum(n=1 to k) 1/[(n + 1)(n + 2)]

= sum(n=1 to k) [1/(n + 1) - 1/(n + 2)], by applying partial fractions

= sum(n=1 to k) 1/(n + 1) - sum(n=1 to k) 1/(n + 2)

= [1/2 + 1/3 + ... + 1/(k + 1)] - [1/3 + 1/4 + ... + 1/(k + 2)]

= 1/2 - 1/(k + 2), as all other terms cancel in pairs.

Letting k --> infinity yields:

sum(n=1 to infinity) 1/[(n + 1)(n + 2)] = 1/2 - 0 = 1/2.

Therefore:

sum(n=1 to infinity) {(1/3)^n - 1/[(n + 1)(n + 2)]}

= sum(n=1 to infinity) (1/3)^n - sum(n=1 to infinity) 1/[(n + 1)(n + 2)]

= 1/2 - 1/2

= 0.

I hope this helps!

∑ n=1 to infinity [(1/3)^n] is a standard geometric series, 1/3 + 1/9 + 1/27 ... = 1/2. (You should have learnt how to do this).

∑ n=1 to infinity [1/((n+1)(n+2))]

= ∑ n=1 to infinity [ ((n+2) - (n+1)) / ((n+1)(n+2))]

= ∑ n=1 to infinity [ 1/(n+1) - 1/(n+2)]

= (1/2 -1/3) + (1/3 - 1/4) + (1/4 - 1/5) ...

= 1/2 + (-1/3 + 1/3) + (- 1/4 + 1/4) ...

= 1/2

So the total is 1/2 - 1/2 = 0

Say you have chose to give up on the kth term, then you definately elect the sum: n.a million+(n-a million).2+(n-2).3+ ... + (n-(ok-a million)).ok = ( n.a million + n.2 + n.3 + ... + n.ok ) - ( a million.2 + 2.3 + 3.4 + ...+ (ok-a million).ok ) = n.ok.(ok+a million) / 2 - ok.(ok-a million).(ok+a million) / 3 = ok.(ok+a million)(3n-2k+2) / 6 If the final term is to be a million.n then you definately might desire to place ok=n, which provides: n(n+a million)(n+2) / 6