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}+...

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asked 2021-03-05
Find a formula for the nth partial sum of each series and use it to find the series’ sum if the series converges.
\(\frac{9}{100}+\frac{9}{100^2}+\frac{9}{100^3}+...+\frac{9}{100^n}+...\)

Answers (1)

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 \(\frac{9}{100}+\frac{9}{100^2}+\frac{9}{100^3}+...+\frac{9}{100^n}+...=\frac{1}{11}\)

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