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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.

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    $\begingroup$ 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, $\frac{1}{1}-\color{red}{\frac{1}{2}} + \color{red}{\frac{1}{2}} - \frac{1}{3} \cdots$ $\endgroup$ – Math Lover Jan 18 '18 at 18:04
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    $\begingroup$ And are you sure about the answer? I think it should be $14/5$ instead. $\endgroup$ – Leo163 Jan 18 '18 at 18:09
  • $\begingroup$ @Leo163 Oops, my bad! That's my dyslexia for you. It's 14/5. $\endgroup$ – numericalorange Jan 18 '18 at 19:00
  • $\begingroup$ @MathLover Yes, I know what those are. Thank you for your insight, this is very useful for me. $\endgroup$ – numericalorange Jan 18 '18 at 19:00
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    $\begingroup$ 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? $\endgroup$ – user21820 Jan 19 '18 at 14:42
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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

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  • $\begingroup$ 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! $\endgroup$ – numericalorange Jan 18 '18 at 19:03
  • $\begingroup$ @numericalorange you are welcome don't forget to vote up it is useful for future users $\endgroup$ – Guy Fsone Jan 18 '18 at 19:07
  • $\begingroup$ 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? $\endgroup$ – numericalorange Jan 18 '18 at 19:13
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    $\begingroup$ @numericalorange that is all thanks $\endgroup$ – Guy Fsone Jan 18 '18 at 19:19
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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 $\frac {14}{3}$

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