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Evaluating

$\sum_{n=1}^\infty \frac{2^{2n-1}}{5^{n+1}}$

Its $\frac{\infty}{\infty}$ so I should use L Hospital rule, but the terms are exponential and differentiation won't do much good? I am thinking maybe I somehow use $\ln$ both sides? But how? Or perhaps I should do something else?

UPDATE

Following @Paul's answer: Since $|r| < 1$ so sequence is convergent. So I use the formula

$\sum_{n=1}^\infty ar^{n-1} = \frac{a}{1-r}$

$\frac{1}{10} \cdot \frac{1}{1-\frac{4}{5}} = \frac{1}{10}\cdot 5 = \frac{1}{2}$

But answer is $\frac{2}{5}$

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    @Gigili: I know, but Jiew Meng either hadn't read or hadn't understood it, so I was reiterating.2012-02-27

2 Answers 2

7

You can rewrite it as $\sum_{n=1}^\infty \frac{2^{2n-1}}{5^{n+1}}=\sum_{n=1}^\infty \frac{\frac{1}{2}\cdot2^{2n}}{5\cdot 5^n}=\frac{1}{10}\sum_{n=1}^\infty \frac{2^{2n}}{5^n}=\frac{1}{10}\sum_{n=1}^\infty \frac{4^{n}}{5^n}=\frac{1}{10}\sum_{n=1}^\infty \left(\frac{4}{5}\right)^n$ which is a geometric series with ratio $r=\displaystyle\frac{4}{5}$. I think you can finish it from here.

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    Sorry, that's a typo. $\frac{1}{10}$ shouldn't appear until the second equality.2012-02-27
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This is a series, not a sequence. L'Hopital's rule is used for the latter, not the former.

When working with series, it is a good idea to first determine if the the given series is of a certain known type. One type is a series of the form $\sum\limits_{n=m}^\infty a r^{n+k} =ar^{m+k} + ar^{m+k+1}++ ar^{m+k+2}+\cdots.$
These are called geometric series and the quantity $r$ is called the ratio of the series.

A geometric series converges if and only if $|r|<1$. When the series converges, it converges to to the first term of the series divided by the quantity $(1-r)$. The first term of the series above is obtained when you take $n=m$: $ar^{m+k}$.

I would not suggest you use a formula to find the sum of the series, but rather do the following:

  1. Identify the ratio $r$.
  2. If $|r|<1$, the series converges to $\text{the first term of the series}\over 1-r $. If $|r|\ge1$, the series diverges.

In your case $ \sum_{n=1}^\infty { {2^{2n-1}\over 5^{n+1}} }=\sum_{n=1}^\infty\textstyle {1\over 10}({4\over5})^n $ This series is geometric with $r=4/5$. The first term of the series is ${1\over10}\cdot{4\over5}$ (obtained by setting $n=1$). So, the series converges and $ \sum_{n=1}^\infty{\textstyle {1\over 10}({4\over5})^n} = {{1\over10}\cdot{4\over5}\over 1-{4\over5}} ={{1\over10}\cdot{4\over5}\over {1\over5}}={2\over5}. $

Another example: $ \sum_{n=2}^\infty 4\cdot (-1/3)^{n+5}\quad \buildrel {r=-1/3}\over{ = }\quad {4\cdot(-1/3)^7\over 1-(-1/3)} ={-4/3^7\over 5/3}. $ Here, the first term of the series is obtained when $n=2$: $4\cdot(-1/3)^{2+5}=4\cdot(-1/3)^{7}$.

And one more: $ \sum_{n=0}^\infty ( 1/3)^{n }\quad \buildrel {r= 1/3}\over{ = }\quad { ( 1/3)^0\over 1-(1/3)} ={1\over 2/3}={3\over2}. $

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    I think the OP just made a mistake while calculating the sum, and Paul linked to the WP article about Geometric series. I once posted an answer completely different from the other answer and I recall you complained why I posted a new answer.2012-02-27