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

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#