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Question # 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}+...

Series
ANSWERED Find a formula for the nth partial sum of each series and use it to find the series’ sum if the series converges.
$$\frac{9}{100}+\frac{9}{100^2}+\frac{9}{100^3}+...+\frac{9}{100^n}+...$$ 2021-03-06

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,
$$s_n=\sum_{k=1}^nb_k$$
$$=\sum_{k=1}^n\frac{9}{100^k}$$
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,
$$s_n=\frac{b(1-r^n)}{1-r}$$
Substitute the values in the above expression,
$$s_n=\frac{(\frac{9}{100})(1-(\frac{1}{100})^n)}{1-\frac{1}{100}}$$
$$=\frac{100}{99}(\frac{9}{100}-\frac{9}{100}(\frac{1}{100})^n)$$
$$=\frac{100}{99}(\frac{9}{100}-9(\frac{1}{100^{n+1}}))$$
Thus the formula for nth partial sum of series is $$=\frac{100}{99}(\frac{9}{100}-9(\frac{1}{100^{n+1}}))$$
The expression for the limit of the nth partial sum of the given series is,
$$L=\lim_{n\rightarrow\infty}s_n$$
$$=\lim_{n\rightarrow\infty}(\frac{100}{99}(\frac{9}{100}-9\lim_{n\rightarrow\infty}(\frac{1}{100^{n+1}})))$$
$$=\frac{100}{99}(\frac{9}{100}-9(\frac{1}{100^{\infty}}))$$
$$=\frac{100}{99}(\frac{9}{100}-9(0))$$
$$=\frac{1}{11}$$
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 $$\frac{9}{100}+\frac{9}{100^2}+\frac{9}{100^3}+...+\frac{9}{100^n}+...=\frac{1}{11}$$