Sum the series:
$$1 + \dfrac13 \cdot \dfrac14 + \dfrac15 \cdot \dfrac1{4^2} + \dfrac17 \cdot \dfrac1{4^3} + \cdots$$

How can I solve this? I am totally stuck on this problem.


Let $$S = \sum_{n=0}^{\infty} \dfrac1{2n+1} \dfrac1{4^n}$$ Note that $$\dfrac1{2n+1} = \int_0^1 x^{2n} dx$$ Hence, we can write \begin{align} S & = \sum_{n=0}^{\infty} \dfrac1{4^n}\int_0^1 x^{2n} dx = \int_0^1 \sum_{n=0}^{\infty} \left(\dfrac{x^2}{4} \right)^n dx\\ & = \int_0^1 \dfrac{dx}{1-\dfrac{x^2}4} = \int_0^1 \dfrac{dx}{2-x} + \int_0^1 \dfrac{dx}{2+x}\\ & = \left. -\log(2-x) \right \vert_0^1 + \left. \log(2+x) \right \vert_0^1 = \log 3 \end{align} Hence, $$\sum_{n=0}^{\infty} \dfrac1{2n+1} \dfrac1{4^n} = \log(3)$$

  • $\begingroup$ swapping a term for an equivalent definite integral is unintuitive at least, replacing a simpler term with a more complex term that requires evaluation should make things harder not easier. But as this example shows the opposite is true, do you know of any more similar type of examples? (i.e. making progress by introducing more complexity ) $\endgroup$ – Arjang Jan 13 '13 at 6:29
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    $\begingroup$ @Arjang There are tons of examples like this. Any sum of the form $\sum_{n=1}^{\infty} P(n) x^n$ where $P(n)$ is a polynomial in $n$ or one over a polynomial in $n$ is amenable to this technique. You may want to look at this problem solved yesterday, where a complex sum is easily solved in terms of an integral. $\endgroup$ – user17762 Jan 13 '13 at 6:37

Using Infinite geometric series, $$1+x+x^2+\cdots=\frac1{1-x}$$ for $|x|<1$

Applying integration wrt to $x, \log(1-x)=-x-\frac{x^2}2-\frac{x^3}3+\cdots+c$ where $c$ is an arbitrary constant of indefinite integral.

Putting $x=0,\log(1)=c\implies c=0$

So, $$\log(1-x)=-x-\frac{x^2}2-\frac{x^3}3+\cdots$$

Putting $x=y$ we get, $$\log(1-y)=-y-\frac{y^2}2-\frac{y^3}3+\cdots$$

Putting $x=-y$ we get, $$\log(1+y)=y-\frac{y^2}2+\frac{y^3}3+\cdots$$

So, $$\log(1+y)-\log(1-y)=2(y+\frac{y^3}3+\frac{y^5}5+\cdots)$$ for $|y|<1$

Consequently, $$1 + \dfrac13 \cdot \dfrac14 + \dfrac15 \cdot \dfrac1{4^2} + \dfrac17 \cdot \dfrac1{4^3} + \cdots=2\left(\frac12+\frac{\left(\frac12\right)^3}3+\frac{\left(\frac12\right)^5}5+\cdots\right)$$ $$=\log(1+\frac12)-\log(1-\frac12)=\log\left(\frac{\frac32}{\frac12}\right)=\log 3$$

  • $\begingroup$ @Arjang, how much is the complexity of this? $\endgroup$ – lab bhattacharjee Jan 13 '13 at 18:16

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