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I'm having trouble understanding infinite sequence and series as it relates to calculus, but I think I'm getting there.

For the below problem:

$$\sum_{n=1}^{\infty}\frac{3}{n(n+3)}$$

The solution shows them breaking this up into a sum of partial fractions. I understand how they got the first two terms, but then they show the partial fractions of the $n$ terms and I find myself lost.

Here is the what I'm talking about:

$$S_n=\sum_{i=1}^{n}\frac{3}{i(i+3)}=\sum_{i=1}^{n}\left(\frac{1}{i}-\frac{1}{i+3} \right)$$

The next few terms are shown to be this:

$$=\left(1-\frac{1}{4}\right)+\left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac{1}{3}-\frac{1}{6}\right)+\left(\frac{1}{4}-\frac{1}{7}\right)+..+$$

And it continues but this is the part where I get confused...

$$\left(\frac{1}{n-3}-\frac{1}{n}\right)+\left(\frac{1}{n-2}-\frac{1}{n+1}\right)+\left(\frac{1}{n-1}-\frac{1}{n+2}\right)+\left(\frac{1}{n}-\frac{1}{n+3}\right)$$

Where did the $n$ terms in the denominator come from?

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  • $\begingroup$ There must be something wrong with the sum as it is written if you don't intend it to be infinite. Perhaps $$ S_n = \sum_{i=1}^n \frac{3}{i(i+3)} $$ $\endgroup$ – Tom Nov 16 '13 at 21:45
  • $\begingroup$ @tom corrected. thank you! $\endgroup$ – hax0r_n_code Nov 16 '13 at 21:53
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    $\begingroup$ No problem. Do you agree with the first few terms? Those $$(1-1/4)+(1/2-1/5)+(1/3-1/6)+(1/4-1/7)+...$$ If so, then what when it continues, it is pointing out that in general (that is if we take a general $n$) and looked at a sequence of four pieces of the sum, it will look like the sum you're confused about. In fact, if you replace $n$ in the general term by $4$, you will see you get the same sum of four terms. $\endgroup$ – Tom Nov 16 '13 at 21:56
  • $\begingroup$ @Tom I see the pattern but why is $4$ special? $\endgroup$ – hax0r_n_code Nov 16 '13 at 22:43
  • $\begingroup$ Actually, your point is exactly why the solution switched to use a general $n$. There is nothing special about 4! This is the general pattern and the case with $n=4$ is just a special case to help you get an intuition for the more general pattern. Good point! $\endgroup$ – Tom Nov 16 '13 at 22:48
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$$\sum_{n=1}^{\infty}\frac{3}{n(n+3)}=\lim_{k\to\infty}\sum_{n=1}^{k}\frac{3}{n(n+3)}=$$ $$=\lim_{k\to\infty}\left(\frac{11}{6}-\left(\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3}\right)\right)=\frac{11}{6}$$ because $$\sum_{n=1}^{k}\frac{3}{n(n+3)}=\sum_{n=1}^{k}\left(\frac{1}{n}-\frac{1}{n+3}\right)=\sum_{n=1}^{k}\frac{1}{n}-\sum_{n=1}^{k}\frac{1}{n+3}=$$ $$=1+\frac{1}{2}+\frac{1}{3}+\sum_{n=4}^{k}\frac{1}{n}-\sum_{n=1}^{k}\frac{1}{n+3}=\frac{11}{6}+\sum_{n=4}^{k}\frac{1}{n}-\sum_{j=4}^{k+3}\frac{1}{j}=$$ $$=\frac{11}{6}+\sum_{n=4}^{k}\frac{1}{n}-\left(\sum_{j=4}^{k}\frac{1}{j}+\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3}\right)=$$ $$=\frac{11}{6}-\left(\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3}\right)$$

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    $\begingroup$ Careful! after the first equality you have $\infty - \infty$ which is undefined. $\endgroup$ – Tom Nov 16 '13 at 21:57
  • $\begingroup$ @Tom. I edit my answer to eliminate $\infty-\infty$ $\endgroup$ – Adi Dani Nov 17 '13 at 0:10
  • $\begingroup$ The partial sums do depend on $k$. The indices on the last sum should run from $j=4$ to $j=k+3$, not $j=4$ to $j=k$. $\endgroup$ – Antonio Vargas Nov 17 '13 at 0:43
  • $\begingroup$ @Antonio Vargas. Thank you very much. I miss that. Now I hope no more mistakes. $\endgroup$ – Adi Dani Nov 17 '13 at 1:14

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