$\displaystyle\sum_{n=0}^{\infty}\frac{1}{(2n+1) . 4^n} $ Can anyone tell me how to approach? It must be convergent as $\displaystyle\sum_{n=0}^{\infty}\frac{1}{4^n}$ is convergent. and answer will be less than 4/3. but how to approach I can not see?
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Can you show that $\frac{1}{2n+1}\cdot\frac{1}{4^n}\leq\frac{1}{4^n}$ and use the comparison test? – Michael Burr Apr 10 '17 at 11:51
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Note that this is the same as $\displaystyle\sum_{n=0}^{\infty}\frac{1}{(2n+1)} (\frac 1 4)^n$, which is a power series where $x = \frac 1 4$ – AsafHaas Apr 10 '17 at 11:51
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Sorry but the tags (ramanujan-summation) and (summation-by-parts) are offtopic here. The tag (summation) is offtopic as well, but the reason is more subtle. Anyway, please check the meaning of the tags before using them. – Did Apr 10 '17 at 11:55
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1See also: Sum of the series $\sum_{n=0}^{\infty} \frac{1}{4^n(2n+1)}$. Found using Approach0. – Martin Sleziak May 28 '17 at 09:00
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Hint. One may write $$ \sum_{n=0}^{\infty}\frac{1}{(2n+1)4^n}=2\sum_{n=0}^{\infty}\int_0^{1/2}t^{2n}dt=2\int_0^{1/2}\sum_{n=0}^{\infty}t^{2n}dt=\int_0^{1/2}\frac2{1-t^2}\:dt $$ the latter may be evaluated by a partial fraction decomposition.
Olivier Oloa
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$$\sum_{n=1}^\infty\dfrac1{(2n+1)4^n}=2\sum_{n=1}^\infty\dfrac{(1/2)^{2n+1}}{2n+1}$$
Now for $1\ge x>-1,$ $$\ln(1+x)=x-\dfrac{x^2}2+\dfrac{x^3}3-\cdots$$
$$\ln(1+x)-\ln(1-x)=?$$
lab bhattacharjee
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See also : http://math.stackexchange.com/questions/365516/why-does-the-taylor-series-of-ln-1-x-only-approximate-it-for-1x-le-1 – lab bhattacharjee Apr 10 '17 at 11:54
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$$\sum_{n=0}^\infty q^n = \frac{1}{1-q}$$ $$\sum_{n=0}^\infty x^{2n} = \frac{1}{1-x^2}$$ $$\int\sum_{n=0}^\infty x^{2n} = \int\frac{1}{1-x^2}$$ $$\sum_{n=0}^\infty \frac{x^{2n+1}}{2n+1} = \tanh^{-1}(x)$$ $$\sum_{n=0}^\infty \frac{x^{2n}}{2n+1} = \frac{\tanh^{-1}(x)}{x}$$ $$\sum_{n=0}^\infty \frac{t^{n}}{2n+1} = \frac{\\tanh^{-1}(\sqrt{t})}{\sqrt{t}}$$ Now evaluate at $t=1/4$
Rab
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