Sum of the infinite series

Question

The series is $$\frac{5}{1\cdot2}\cdot\frac{1}{3}+\frac{7}{2\cdot3}\cdot\frac{1}{3^2}+\frac{9}{3\cdot4}\cdot\frac{1}{3^3}+\frac{11}{4\cdot5}\cdot\frac{1}{3^4}+\cdots$$ This is my attempt: $$T_n=\frac{2n+3}{n(n+1)}\cdot\frac{1}{3^n}$$ Assuming $$\frac{2n+3}{n(n+1)}=\frac{A}{n}+\frac{B}{n+1}$$ we find $A=3,B=-1.$ Putting these values in $T_n$ we get, $$T_n=\frac{1}{n}\cdot\frac{1}{3^{n-1}}-\frac{1}{n+1}\cdot\frac{1}{3^{n}}$$ How do I find the sum of the series from here $?$

Writing $T_n=3/n3^n-3/(n+1)3^{n+1}$ makes the telescoping more obvious. — , Jan 04 '18 at 10:36

jonsno · Accepted Answer · 2018-01-03T14:33:54.080

4

Just try to write the expression for $T_{n+1}$ and note that the series telescopes.

$$T_n=\frac{1}{n}\cdot\frac{1}{3^{n-1}}-\color{purple}{\frac{1}{n+1}\cdot\frac{1}{3^{n}}} \\ T_{n+1}=\color{purple}{\frac{1}{n+1}\cdot\frac{1}{3^{n}}}-\frac{1}{n+2}\cdot\frac{1}{3^{n+1}}$$

We see the terms are cancelling, so

$$\begin{align} \sum_{i=1}^{n} T_i &= \frac{1}{1}\frac{1}{3^{0}} -\color{blue}{\frac{1}{2}\frac{1}{3^{1}}+\frac{1}{2}\frac{1}{3^{1}}}... -\color{red}{\frac{1}{n-1}\frac{1}{3^{n-2}}+\frac{1}{n-1}\frac{1}{3^{n-2}}} -\frac{1}{n}\frac{1}{3^{n-1}}\\ &= 1-\frac{1}{n3^{n-1}} \end{align}$$

Therefore as $n\to \infty$, we see the sum tends to $1$.

edited Jan 03 '18 at 14:33

answered Jan 03 '18 at 14:02

jonsno

7,521

I don't see how this will help to conclude. – Olivier Oloa Jan 03 '18 at 14:04
@OlivierOloa We add $T_n$ with $T_{n+1}$ and terms cancel ? – jonsno Jan 03 '18 at 14:05
We are looking for a closed form of $\sum T_n$. – Olivier Oloa Jan 03 '18 at 14:07
@OlivierOloa I am sorry I still don't see what you are referring to! I followed where OP left and see if we add, all terms cancel and only the first term remains, ie $\frac{1}{1}\cdot \frac{1}{3^{1-1}} = 1$. am I missing something? – jonsno Jan 03 '18 at 14:10
Sorry I don't understand how you obtain a final result from your two lines above. – Olivier Oloa Jan 03 '18 at 14:12
@OlivierOloa See in $T_{n} + T_{n+1}$ the purple terms cancel. Then if we add $T_{n+2}$ the next term $\frac{1}{n+2}.\frac{1}{3^{n+1}}$ cancels. Similarly all terms cancel and as $n\to \infty$ the last term becomes $0$. Ill better edit it in.. – jonsno Jan 03 '18 at 14:15
Let us continue this discussion in chat. – jonsno Jan 03 '18 at 14:24
Oh I see it now, thanks! (+1) – Olivier Oloa Jan 03 '18 at 14:33
Ok cheers :) This was most straightforward as OP had done all the work! There must be other interesting methods! – jonsno Jan 03 '18 at 14:36
Telescoping terms is often the simplest method! – Olivier Oloa Jan 03 '18 at 14:37

Peter Szilas · Answer 2 · 2018-01-04T10:40:21.820

$\sum_{i=1}^{n} T_i =S_n - R_n$ , where

$S_n := \sum_{i=1}^{n} \dfrac{1}{i}\dfrac{1}{3^{i-1}};$

$R_n :=\sum_{i=1}^{n} \dfrac{1}{i+1} \dfrac{1}{3^i}.$

Change the dummy index in $R_n:$

$k= i +1$, then

$R_n=\sum_{k=2}^{n+1} \dfrac{1}{k}\dfrac{1}{3^{k-1}} =$

$(1 + \sum_{k=2}^{n}\dfrac{1}{k}\dfrac{1}{3^{k-1}}) -1 + \dfrac{1}{n+1}\dfrac{1}{3^n}$

$= S_n - 1 + \dfrac{1}{n+1}\dfrac{1}{3^n}.$

Hence:

$T_n = S_n - R_n= 1 - \dfrac{1}{ n+1}\dfrac{1}{3^n}.$

score 0 · Answer 3 · answered Jan 03 '18 at 19:31

$$Sum=\sum_{n=1}^{\infty} \frac{1}{n}.\frac{1}{3^{n-1}}-\frac{1}{n+1}.\frac{1}{3^{n}}$$

$$Sum=\sum_{n=1}^{\infty} \frac{1}{n}.\frac{1}{3^{n-1}}-\sum_{n=1}^{\infty} \frac{1}{n+1}.\frac{1}{3^{n}}$$

$$Sum=\sum_{n=1}^{\infty} \frac{1}{n}.\frac{1}{3^{n-1}}-\sum_{n=2}^{\infty} \frac{1}{n}.\frac{1}{3^{n-1}}$$

$$Sum=\sum_{n=1}^{1} \frac{1}{n}.\frac{1}{3^{n-1}}=1$$

Sum of the infinite series

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