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

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?


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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L'Hopital's rule is for evaluating limits, not evaluating series. – JavaMan Feb 27 '12 at 13:00
What's the point of your update? If you follow Paul's answer, you'll have it. The first term is "$\frac{4}{5}$", not $1$. – Gigili Feb 27 '12 at 13:11
@Jiew Meng: Compare the series $\sum_{n=1}^\infty \left(\frac{4}{5}\right)^n=\sum_{n=1}^\infty \frac{4}{5}\left(\frac{4}{5}\right)^{n-1}$ with $\sum_{n=1}^\infty ar^{n-1}$, we have $a=\frac{4}{5}$ and $r=\frac{4}{5}$. Therefore, using the formula you have, we have $\sum_{n=1}^\infty \left(\frac{4}{5}\right)^n=\frac{\frac{4}{5}}{1-\frac{4}{5}}=4$. – Paul Feb 27 '12 at 13:16
@CliveNewstead: That's what I said! – Gigili Feb 27 '12 at 13:18
@Gigili: I know, but Jiew Meng either hadn't read or hadn't understood it, so I was reiterating. – Clive Newstead Feb 27 '12 at 14:23
up vote 7 down vote accepted

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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I think there is a $\frac{1}{10}$ too in the first step. – Riccardo.Alestra Feb 27 '12 at 12:36
Actually shouldn't the $\frac{1}{10}$ appear in step 3? – Jiew Meng Feb 27 '12 at 12:44
Btw, I also updated my question – Jiew Meng Feb 27 '12 at 12:58
Sorry, that's a typo. $\frac{1}{10}$ shouldn't appear until the second equality. – Paul Feb 27 '12 at 13:13

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 don't see what special points you added that Paul didn't mention in his answer or it's not mentioned in comments. – Gigili Feb 27 '12 at 13:56
@Gigili Mostly how to actually find the sum of a geometric series. It seemed to OP had problems using the formula. – David Mitra Feb 27 '12 at 13:59
@Gigli Oh, I didn't see the hidden comments... – David Mitra Feb 27 '12 at 14:00
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. – Gigili Feb 27 '12 at 14:04

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