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Q 50.

Expert-verifiedFound in: Page 615

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

Determine whether the series $\sum _{n=4}^{\infty}\frac{4}{{11}^{n}}$ converges or diverges. Give the sum of the convergent series.

The series $\sum _{n=4}^{\infty}\frac{4}{{11}^{n}}$ converges to $\frac{2}{6655}$.

Given a series $\sum _{n=4}^{\infty}\frac{4}{{11}^{n}}$.

The index starts with 2, rather than 0.

Note that the convergence of a series depends not upon the first few terms but only upon the tail of the series.

The standard form of a geometric series is $\sum _{k=0}^{\infty}c{r}^{k}$.

The geometric series converges if and only if $\left|r\right|<1$.

Here, the series $\sum _{n=4}^{\infty}\frac{4}{{11}^{n}}$ has $c=\frac{4}{{11}^{4}}$ and $r=\frac{1}{11}$.

Since $r=\frac{1}{11}$, it follows that the series localid="1648886017183" $\sum _{n=4}^{\infty}\frac{4}{{11}^{n}}$ converges.

If the geometric series $\sum _{k=0}^{\infty}c{r}^{k}$ converges, it converges to $\frac{c}{1-r}$.

So, the series $\sum _{n=4}^{\infty}\frac{4}{{11}^{n}}$ converges to $\frac{\frac{4}{{11}^{4}}}{1-\frac{1}{11}}$, that is $\frac{4}{1331}$ or equivalently $\frac{2}{6655}$.

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