# Find a formula for the nth partial sum of each series and use it to find the series’ sum if the series converges. frac{9}{100}+frac{9}{100^2}+frac{9}{100^3}+...+frac{9}{100^n}+...

Find a formula for the nth partial sum of each series and use it to find the series’ sum if the series converges.
$\frac{9}{100}+\frac{9}{{100}^{2}}+\frac{9}{{100}^{3}}+...+\frac{9}{{100}^{n}}+...$
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saiyansruleA

Given data
The series is: $\frac{9}{100}+\frac{9}{{100}^{2}}+\frac{9}{{100}^{3}}+...+\frac{9}{{100}^{n}}+...$
Consider the infinite series and compare with it given series,
${b}_{1}+{b}_{2}+{b}_{3}+...+{b}_{n}+...$
${b}_{n}=\frac{9}{{100}^{n}}$
The nth partial sum of the given series is,
${s}_{n}=\sum _{k=1}^{n}{b}_{k}$
$=\sum _{k=1}^{n}\frac{9}{{100}^{k}}$
The first term of the given series is: $b=\frac{9}{100}$
The common ratio is: $r=\frac{1}{100}$
The expression for the sum of series is,
${s}_{n}=\frac{b\left(1-{r}^{n}\right)}{1-r}$
Substitute the values in the above expression,
${s}_{n}=\frac{\left(\frac{9}{100}\right)\left(1-\left(\frac{1}{100}{\right)}^{n}\right)}{1-\frac{1}{100}}$
$=\frac{100}{99}\left(\frac{9}{100}-\frac{9}{100}\left(\frac{1}{100}{\right)}^{n}\right)$
$=\frac{100}{99}\left(\frac{9}{100}-9\left(\frac{1}{{100}^{n+1}}\right)\right)$
Thus the formula for nth partial sum of series is $=\frac{100}{99}\left(\frac{9}{100}-9\left(\frac{1}{{100}^{n+1}}\right)\right)$
The expression for the limit of the nth partial sum of the given series is,
$L=\underset{n\to \mathrm{\infty }}{lim}{s}_{n}$
$=\underset{n\to \mathrm{\infty }}{lim}\left(\frac{100}{99}\left(\frac{9}{100}-9\underset{n\to \mathrm{\infty }}{lim}\left(\frac{1}{{100}^{n+1}}\right)\right)\right)$
$=\frac{100}{99}\left(\frac{9}{100}-9\left(\frac{1}{{100}^{\mathrm{\infty }}}\right)\right)$
$=\frac{100}{99}\left(\frac{9}{100}-9\left(0\right)\right)$
$=\frac{1}{11}$
The nth partial sum of the given series is $\frac{1}{11}$, and the sequence of the partial sum converges to limit $L=\frac{1}{11}$, so the given series converges.
Thus the sum of the series is $\frac{9}{100}+\frac{9}{{100}^{2}}+\frac{9}{{100}^{3}}+...+\frac{9}{{100}^{n}}+...=\frac{1}{11}$

Jeffrey Jordon