Calculate the sum of the series: $\sum\limits_{n=1}^{\infty}\left ( \frac{3}{n^{2}+n}-\frac{2}{4n^{2}+16n+15} \right )$

I am stuck at a part when it comes to evaluating the sum of the said series... Here is my work so far (and I am not sure if the notation and simplification is correct either):

Simplifying using partial sums:

$$\sum_{n=1}^{\infty}\left ( \frac{3}{n^{2}+n}-\frac{2}{4n^{2}+16n+15} \right )=\sum_{n=1}^{\infty}\left ( 3\left ( \frac{1}{n}-\frac{1}{n+1} \right )-\left ( \frac{1}{2n+3}-\frac{1}{2n+5} \right ) \right )$$

Now I take the limit of the Nth partial sum of the series, right?

$$\lim_{N\rightarrow \infty}\sum_{n=1}^{N}\left ( 3\left ( \frac{1}{N}-\frac{1}{N+1} \right )-\left ( \frac{1}{2N+3}-\frac{1}{2N+5} \right ) \right )$$

$$=3\lim_{N\rightarrow \infty}\sum_{n=1}^{N}\left ( \frac{1}{N}-\frac{1}{N+1} \right )-\lim_{N\rightarrow \infty}\sum_{n=1}^{N}\left ( \frac{1}{2N+3}-\frac{1}{2N+5} \right )$$

$$=3\lim_{N\rightarrow \infty}\sum_{n=1}^{N} \frac{1}{N}-3\lim_{N\rightarrow \infty}\sum_{n=1}^{N}\frac{1}{N+1} -\lim_{N\rightarrow \infty}\sum_{n=1}^{N} \frac{1}{2N+3}+\lim_{N\rightarrow \infty}\sum_{n=1}^{N}\frac{1}{2N+5}$$

I assume I doing something wrong here, because each term diverges. This is what was written in the textbook:

$$\sum_{n=1}^{\infty}\left ( \frac{1}{n}-\frac{1}{n+1} \right )=1-\lim_{N\rightarrow \infty}\frac{1}{N}=1$$

$$\sum_{n=1}^{\infty}\left ( \frac{1}{2n+3}-\frac{1}{2n+5} \right )=\frac{1}{5}-\lim_{N\rightarrow \infty}\frac{1}{2N+3}=\frac{1}{5}$$

The problem is...I am not sure how they got this! They are missing a lot of steps for me to understand, hence the messiness above.

• Are you aware of telescopic sums? If not, then just write first few terms of $\frac{1}{n}-\frac{1}{n+1}$ and see what happens when we sum them. For example, $\frac{1}{1}-\color{red}{\frac{1}{2}} + \color{red}{\frac{1}{2}} - \frac{1}{3} \cdots$ Jan 18, 2018 at 18:04
• And are you sure about the answer? I think it should be $14/5$ instead. Jan 18, 2018 at 18:09
• @Leo163 Oops, my bad! That's my dyslexia for you. It's 14/5.
– user482939
Jan 18, 2018 at 19:00
• @MathLover Yes, I know what those are. Thank you for your insight, this is very useful for me.
– user482939
Jan 18, 2018 at 19:00
• I'm not sure why nobody has pointed out so far that your error is in making an unsound deduction. The limit of a sum/difference is not necessarily the sum/difference of the individual terms' limits. Observe that $\lim_{n\to\infty} (n-n) = 0$ but $\lim_{n\to\infty} n$ simply does not exist. You can interchange the limit and the finite sum/difference if the individual terms' limits exist. If not, then no go. Always remember that if you get nonsensical results, you must have made an unsound deduction somewhere. Makes sense? Jan 19, 2018 at 14:42

Note that both series $$\sum_{n=1}^{\infty}\frac{3}{n^{2}+n}~~and ~~~~\sum_{n=1}^{\infty}\frac{2}{4n^{2}+16n+15}$$

converges then there is no risk to separate the sum.

However using telescoping sum for the second sum as follows let $u_n=\frac{1}{2n+3}$ then $u_{n+1}=\frac{1}{2n+5}$ hence

$$~~~~\sum_{n=1}^{\infty}\frac{2}{4n^{2}+16n+15} =\lim_{k\to\infty }\sum_{n=1}^{k}\left ( \frac{1}{2n+3}-\frac{1}{2n+5} \right )\\\color{red}{:=\lim_{k\to\infty }\sum_{n=1}^{k}\left ( u_n-u_{n+1} \right )=\lim_{k\to\infty }(u_1-u_{k+1})}\\=\frac{1}{5}-\lim_{k\rightarrow \infty}\frac{1}{2k+5}=\frac{1}{5}$$

whereas the first is obvious since $$\sum_{n=1}^{\infty}\frac{3}{n^{2}+n} = 3\lim_{k\to\infty }\sum_{n=1}^{k}\left(\frac{1}{n}-\frac{1}{n+1}\right) =3\lim_{k\to\infty }(1-\frac{1}{k+1}) = 3~~$$

and

• I see now. This is based off the telescoping series, and the fact that the first and second term are of this form. Thanks so much!
– user482939
Jan 18, 2018 at 19:03
• @numericalorange you are welcome don't forget to vote up it is useful for future users Jan 18, 2018 at 19:07
• I upvoted you and pressed the green check mark to show this is the best answer. Is that what you mean by upvoting it? Or do I upvote my own question?
– user482939
Jan 18, 2018 at 19:13
• @numericalorange that is all thanks Jan 18, 2018 at 19:19

Hint:

The given expression can be written as $$\sum_{n=1}^\infty \frac {3}{n(n+1)} - \frac {2}{(2n+3)(2n+5)}$$

On partial decomposition it becomes

$$\sum_{n=1}^\infty \left[3\left(\frac {1}{n}-\frac {1}{n+1}\right) - \left(\frac {1}{2n+3}-\frac{1}{2n+5}\right) \right]$$

Can you see the series telescoping. By the way the answer I guess might be $\frac {14}{3}$