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Here in example 4 this result is shown.

${s_n} = \sum\limits_{i = 1}^n {\dfrac{1}{{{3^{i - 1}}}}} = \dfrac{3}{2}\left( {1 - \dfrac{1}{{{3^n}}}} \right)$

Now without looking at its solution I was trying like this:

${s_n} = \sum\limits_{i = 1}^n {\dfrac{1}{{{3^{i - 1}}}}}=3\sum\limits_{i = 0}^n {\dfrac{1}{{{3^{i }}}}}=3\bigg(\dfrac{1-(\frac{1}{3})^{n+1}}{1-\frac{1}{3}}\bigg)=\mathop {\lim }\limits_{n \to \infty }\dfrac{3}{2}\cdot3(1-(\frac{1}{3})^{n+1})=\dfrac{9}{4}$

What am I doing wrong why I am not getting answer $\dfrac{3}{2}$

Daman
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    Why are you taking a limit as $n \to \infty$? Also, the additional factor of $3$ you introduce in the second step is incorrect. We just have $\sum_{i=1}^n \frac{1}{3^{i-1}} = \sum_{i=0}^{n-1} \frac{1}{3^i}$ (note the $n-1$), as is the $4$ in the final step. – Qiaochu Yuan Jan 05 '21 at 07:18
  • @QiaochuYuan I want to check if given series is convergent or divergent. – Daman Jan 05 '21 at 07:19
  • @Daman: It’s the sum of finitely many terms: convergence isn’t an issue. – Brian M. Scott Jan 05 '21 at 07:19
  • @QiaochuYuan isnt it $\frac{1}{3^{t-1}}$ is $\frac{3}{3^t}$ thats why I took 3 out – Daman Jan 05 '21 at 07:20
  • @Daman: Just shift the index: $$\sum_{i=1}^n\frac1{3^{i-1}}=\sum_{i=0}^{n-1}\left(\frac13\right)^i=\frac{1-\frac1{3^n}}{1-\frac13}=\frac32\left(1-\frac1{3^n}\right)$$ Note that you converted a sum of $n$ terms into a sum of $n+1$ terms; that should have made you suspicious. – Brian M. Scott Jan 05 '21 at 07:21
  • In that case it's the lower index of $0$ which is incorrect; we have $\sum_{i=1}^n \frac{1}{3^{i-1}} = 3 \sum_{i=1}^n \frac{1}{3^i}$ (the indices of summation are unchanged). – Qiaochu Yuan Jan 05 '21 at 07:21
  • @QiaochuYuan: Thanks; even caught it in time to fix it! – Brian M. Scott Jan 05 '21 at 07:22
  • @QiaochuYuan Tell me one thing when it is to decide I have to start my partial sum from 0 or 1 ? Could you guys please see the link and tell me. I am confused when it has to start 1 or 0. – Daman Jan 05 '21 at 07:23
  • @Daman: You were given a series starting at $n=1$, so your partial sum of the first $n$ terms will be exactly what you had: $\sum_{i=1}^n\frac1{3^{i-1}}$. – Brian M. Scott Jan 05 '21 at 07:25
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    You can start the sum wherever you want as long as you know how to evaluate it. If you want to invoke the sum of a geometric series in the form $\sum_{i=0}^{n-1} r^i = \frac{1 - r^n}{1 - r}$ then the sum has to start at $0$, but wherever it starts you can factor out the first term so it doesn't really matter where you start it as long as you're careful not to make an algebra mistake when you shift the indices around and so forth (you've made two). – Qiaochu Yuan Jan 05 '21 at 07:26
  • @QiaochuYuan Thanks for the guidance. – Daman Jan 05 '21 at 07:59

1 Answers1

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$S_n=\sum_{i=1}^n \frac{1}{3^{i-1}}=1+\frac 13+...+\frac{1}{3^{n-1}}=\frac{1-\frac{1}{3^n}}{1-\frac13}=\frac 32(1-\frac{1}{3^n})$ Thus $\lim_{n \to\infty}S_n=\frac 32$. If you are considering the infinite geometric series with $|r|<1$ then sum=$\frac{a}{1-r}$ where $a$ is the first term of the series. Here $a=1,r=\frac 13$. So sum is $\frac{1}{1-\frac 13}=\frac 32$

Saikai Prime
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