# Finding the summation of these two Infinite Series

I find the sequence of partial sums, but I cannot find the general term. I know once I find the general term of the sequence of partial sums, I take the limit of that and that number is the sum of the series. I just cannot find a pattern in either of these.

Your help would be greatly appreciated.

• Hints : Partial fraction decomposition and geometric series – Peter Mar 27 '18 at 20:26
• The first can be written as a telescoping series. The second can be writen as the sum of two distinct geometric series. Can you wrap it up? – Mark Viola Mar 27 '18 at 20:41
• Thank you for your hints, it's so easy once you know how to do it! – user357335 Mar 27 '18 at 21:01
• @user357335 Please remember that you can choose an answer among the given if the OP is solved, more details here meta.stackexchange.com/questions/5234/… – gimusi Apr 1 '18 at 7:47

• $\sum \frac{2}{n^2+4n+3}=\sum \frac{1}{n+1}-\sum\frac{1}{n+3}$ and many terms cancel out
• $\sum \frac{1+2^n}{3^n}=\sum \left(\frac13\right)^n+\sum \left(\frac23\right)^n$ and refer to geometric series
To elaborate on the comment hint, write the first sum as $$\sum \left(\frac{A}{n+B} - \frac{C}{n+D}\right)$$ and apply telescoping series. For the second one note that $$\frac{1+2^n}{3^n} = \frac{1}{3^n} + \frac{2^n}{3^n} = (1/3)^n + (2/3)^n$$