Question 328453
This is kind of tricky trying to set up a general form for 5,55,555,....

We can think of the numerators: 5,55,555,5555,...., as

{{{(5/9)(10^n-1)}}}

But, this gives us a general form of

{{{(5/9)*sum((10^n-1)/13^n,n=1,infinity)}}}

{{{5/9*(sum((10/13)^n,n=1,infinity)-sum((1/13)^n,n=1,infinity))}}}

Using the formula for the sum of an infinite series we get 

{{{(10/13)/(1-10/13)=10/3}}}

{{{(1/13)/(1-1/13)=1/12}}}

{{{5/9*(10/3-1/12)=65/36}}}

The sum is 65/36