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I have got the following summation-$$\displaystyle\sum_{r=1}^{n}\frac{r^24^r}{(r+1)(r+2)}.$$

I have to find the closed form or the general form to find the sum of this series. I know upto summation of Telescopic Series and Some special series like those in $AP$ or $GP$.

But I have no idea on how to begin on this problem.

Thanks for any help!!

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  • $\begingroup$ Do you know calculus? $\endgroup$ – Mark Viola Nov 26 '16 at 20:19
  • $\begingroup$ @Dr.MV Yeah....beginner actually.... $\endgroup$ – tatan Nov 27 '16 at 11:24
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Since you requested a more detailed answer than that of @N.S, here you go.

We start with the series you give, $$\displaystyle\sum_{r=1}^{n}\frac{r^24^r}{(r+1)(r+2)}$$ We now want to simplify this into a series you can deal with; the first step will be to run partial fraction decomposition on the term $$\frac{r^2}{(r+1)(r+2)}=\frac{r^2+3r+2-3r-2}{(r+1)(r+2)}=\frac{(r+1)(r+2)-3r-2}{(r+1)(r+2)}$$ $$=1-\frac{3r+2}{(r+1)(r+2)} =1+\frac{1}{r+1}-\frac{4}{r+2}$$ Let me know if you need help with this partial fraction decomposition. We can now rewrite the above series as $$\displaystyle\sum_{r=1}^{n}\left(4^r+\frac{4^r}{r+1}-\frac{4^{r+1}}{r+2}\right)=\displaystyle\sum_{r=1}^{n}\left(4^r\right)+\sum_{r=1}^{n}\left(\frac{4^r}{r+1}-\frac{4^{r+1}}{r+2}\right)\tag{1}$$ The first series is geometric, which you say you know how to evaluate. (It comes out to be $\frac{4}{3}(4^n - 1)$)

To evaluate the second series in $(1)$ we will look at the expanded terms: $$\sum_{r=1}^{n}\left(\frac{4^r}{r+1}-\frac{4^{r+1}}{r+2}\right)$$ $$= \left(\frac{4}{2}\color{red}{-\frac{4^2}{3}}\right)+\left(\color{red}{\frac{4^2}{3}}\color{green}{-\frac{4^3}{4}}\right)+\left(\color{green}{\frac{4^3}{4}}\color{purple}{-\frac{4^4}{5}}\right)+\cdots+\left(\color{olive}{\frac{4^{n-1}}{n}}\color{fuchsia}{-\frac{4^n}{n+1}}\right)+\left(\color{fuchsia}{\frac{4^n}{n+1}}\color{navy}{-\frac{4^{n+1}}{n+2}}\right)$$ As my above, colored terms hopefully make clear almost all the terms will cancel out with their neighboring term except for the first and last terms (this is called a telescoping series) and so this series comes out to be $$\sum_{r=1}^{n}\left(\frac{4^r}{r+1}-\frac{4^{r+1}}{r+2}\right) = \frac{4}{2}-\frac{4^{n+1}}{n+2}=2-\frac{4^{n+1}}{n+2}$$ We can now go plug this back into $(1)$ and rewrite your original series as $$\frac{4}{3}(4^n - 1)+2-\frac{4^{n+1}}{n+2}= 4^{n+1}\left(\frac{1}{3}-\frac{1}{n+2}\right)+\frac{2}{3}$$ Note that, as $n\to\infty$, this expression becomes closer and closer to being $$4^{n+1}\left(\frac{1}{3}-0\right)+\frac{2}{3}=\frac{4^{n+1}+2}{3}$$ And so the limit towards $+\infty$ clearly diverges to $+\infty$; likewise, the limit towards $-\infty$ diverges towards $-\infty$.

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  • $\begingroup$ Awesome....You deserve the (+50) bounty points!! $\endgroup$ – tatan Nov 27 '16 at 11:38
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    $\begingroup$ Can you please be more explanatory with the partial fraction part?Why did you break it up into the $+3r+2$ part....what was the intuition behind it? $\endgroup$ – tatan Nov 27 '16 at 11:39
  • $\begingroup$ @tatan in short? I read N.S.'s answer and noticed how to get from step one to step two in his post without doing polynomial long division (because he had already done it for me! I often use an online tool like WolframAlpha to give me the answer then I find a quicker way to get there like the trick I showed). In essence, I wanted to make the highest power of $r$ in the numerator be one, because if we expand the denominator we get a term $r^2$ and you want the numerator to be lower degree in order to use partial fraction decomposition $\endgroup$ – Brevan Ellefsen Nov 27 '16 at 16:18
  • $\begingroup$ @tatan if you want, I can put up a tutorial for either partial fraction decomposition, polynomial long division, or both. If you were to do this problem entirely by hand you would have to use both! $\endgroup$ – Brevan Ellefsen Nov 27 '16 at 16:19
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Hint $$\frac{r^2}{(r+1)(r+2)}=1-\frac{3r+2}{(r+1)(r+2)}=1+\frac{1}{r+1}-\frac{4}{r+2}$$

Hint 2 $$4^r\frac{4}{r+2}=\frac{4^{r+1}}{r+2}$$

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