Question

How to calculate this limit?`lim_(n->oo)sum_(k=1)^n(1/2^k+1/3^k)`

Answer 1

`lim_(n->oo) sum_(k=1)^n (1/2^k + 1/3^k) = 3/2`

Explanation:

The general term of a geometric series can be written:

`a_k = a*r^(k-1)`

where `a` is the initial term and `r` the common ratio.

Then we find:

`(1-r) sum_(k=1)^n a_k = sum_(k=1)^n a*r^(n-1) - r sum_(k=1)^n a*r^(n-1)`

`color(white)((1-r) sum_(k=1)^n a_n) = sum_(k=1)^n a*r^(n-1) - sum_(k=2)^(n+1) a*r^(n-1)`

`color(white)((1-r) sum_(k=1)^n a_n) = a+color(red)(cancel(color(black)(sum_(k=2)^n a*r^(n-1)))) - color(red)(cancel(color(black)(sum_(k=2)^n a*r^(n-1)))) - a*r^n`

`color(white)((1-r) sum_(k=1)^n a_k) = a(1 - r^n)`

Divide both ends by `(1-r)` to find:

`sum_(k=1)^n a_k = (a(1-r^n))/(1-r)`

Both `1/2^k` and `1/3^k` are the terms of geometric sequences, the former with initial term `1/2` and common ratio `1/2`, the latter with initial term `1/3` and common ratio `1/3`.

So we find:

`sum_(k=1)^n (1/2^k + 1/3^k) = sum_(k=1)^n 1/2^k + sum_(k=1)^n 1/3^k`

`color(white)(sum_(k=1)^n (1/2^k + 1/3^k)) = (1/2(1-(1/2)^n))/(1-1/2) + (1/3(1-(1/3)^n))/(1-1/3)`

`color(white)(sum_(k=1)^n (1/2^k + 1/3^k)) = (1-(1/2)^n) + 1/2(1-(1/3)^n)`

Then:

`lim_(n->oo) sum_(k=1)^n (1/2^k + 1/3^k) = lim_(n->oo) ((1-(1/2)^n) + 1/2(1-(1/3)^n))`

`color(white)(lim_(n->oo) sum_(k=1)^n (1/2^k + 1/3^k)) = 1+1/2`

`color(white)(lim_(n->oo) sum_(k=1)^n (1/2^k + 1/3^k)) = 3/2`