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Limits — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSLimits3 Marks
Question
Evaluate the following limit:
$\lim\limits_{\text{n}\rightarrow\infty}\Big(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\ \cdots+\frac{1}{3^\text{n}}\Big)$
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Answer

$\lim\limits_{\text{n}\rightarrow\infty}\Big(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\ \cdots+\frac{1}{3^\text{n}}\Big)\ \cdots(\text{i})$
This is G.P. of common ratio $\frac13$
$\therefore$ Sum of n term of G.P. with $\text{a} = \frac13,\text{r}=\frac{-1}{3}$
$\text{S}_\text{n}=\text{a}\Big(\frac{1-\text{r}^{\text{n}}}{1-\text{r}}\Big)$
$\text{S}_\text{n}=\frac13\Bigg(\frac{1-\big(\frac{1}{3}\big)^\text{n}}{1-\frac13}\Bigg)$
$=\frac13\Bigg(\frac{1-\frac{1}{3^\text{n}}}{\frac23}\Bigg)$
$=\frac13\times\frac32\Big(1-\frac{1}{3^\text{n}}\Big)$
$\text{S}_\text{n}=\frac12\Big(1-\frac{1}{3^\text{n}}\Big)$
Substituting value of Sn in (i), we get
$\lim\limits_{\text{n}\rightarrow\infty}​​\text{S}_\text{n}=\lim\limits_{\text{n}\rightarrow\infty}​​\frac12\Big(1-\frac{1}{3^\text{n}}\Big)$
$=\frac{1}{2}\lim\limits_{\text{n}\rightarrow\infty}​​\Big(1-\frac{1}{3^\text{n}}\Big)$
$=\frac12(1-0)$
$=\frac12$

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