# Summation of this series ${1\over 2}\left[1+{1 \over 3}\left({1 \over 4}\right)^2+ {1 \over 5}\left({1 \over 4}\right)^4+\cdots\right]$

I have this series,

$${1\over 2}\left[1+{1 \over 3}\left({1 \over 4}\right)^2+ {1 \over 5}\left({1 \over 4}\right)^4+\cdots\right]$$

I have shown that this series converges.

It looks like some combination of geometric series and harmonic series to me.

I am not able to proceed from here. I would love to have some hint.

Sorry this time I have nothing to write in "my efforts" as I am completely stuck.

Hint:

$$\sum_{n=0}^{\infty}\frac{x^{2n}}{2n+1}=\frac{1}{x}\int_{0}^{x}\left(\sum_{n=0}^{\infty}u^{2n}\right)du$$

Can you take it from here?

Some more explanations:

Notice that each term in your sum can be written like $$\frac{x^{2n}}{2n+1}\tag{1}$$ where $x=1/4$. So your sum can be written: $$\frac{1}{2}\sum_{n=0}^{\infty}\frac{x^{2n}}{2n+1}\tag{2}$$Now we know that the anti-derivative of $x^{2n}$ is $\frac{x^{2n+1}}{2n+1}$, and this looks very much like $\frac{x^{2n}}{2n+1}$, except that there is an $x$ too much!

We can fix this problem by taking the anti-derivative, and then divide by $x$. This would look like $$\frac{x^{2n}}{2n+1}=\frac{1}{x}\int_{0}^{x}u^{2n}du=\frac{1}{x}\left(\frac{x^{2n+1}}{2n+1}-\frac{0^{2n+1}}{2n+1}\right)\tag{3}$$ for each term. Now you can just take the sum of both sides, and you should get your original sum.

You can now write you sum as an integral, with integrand $$\sum_{n=0}^{\infty}u^{2n}\tag{4}$$ can you simplify this?

Just to finish:

You end up with the integral in the hint, and the sum $(4)$ is a geometric series which converges to $$\frac{1}{1-u^2}=\frac{1}{2}\left(\frac{1}{1-u}+\frac{1}{1+u}\right)\tag{5}$$ We can now compute the integral: $$\frac{1}{4x}\int_{0}^{x}\frac{1}{1-u}+\frac{1}{1+u}du=\frac{1}{4x}\ln\left(\frac{1+x}{1-x}\right)\tag{6}$$ Setting $x=1/4$ yields: $$\ln\left(\frac{5}{3}\right)\approx 0.51$$

• Give me 5 minutes, I will let you know. – StammeringMathematician Sep 11 '18 at 13:49
• How did you get this inequality and there is $x$ on LHS and $u$ on RHS, I am not quite able to understand the idea. Can you give me some more hint? – StammeringMathematician Sep 11 '18 at 13:59
• @StammeringMathematician I've edited my answer, please tell me if it helps. – cansomeonehelpmeout Sep 11 '18 at 14:09
• I got it thanks. – StammeringMathematician Sep 11 '18 at 14:17
• @StammeringMathematician real-analysis is not my field, so I'm not comfortable recommending books, sadly. The more general result is linked to Fubini's theorem. Here is a better answer. Good luck! :) – cansomeonehelpmeout Sep 11 '18 at 15:07

Hint:You may use following function:

$$S=1+\frac{1}{3}x^2+\frac{1}{5}x^4+ . . .$$

Multiplying both sides by x we get:

$$S.x=x+\frac{1}{3}x^3+\frac{1}{5}x^5+ . . .$$

Taking derivative we get:

$$S'x+ S=1+x^2+x^4 +. . . (x^2)^n=\frac{(x^2)^{n+1} -1}{x^2-1}$$

Taking integral we get:

$$S.x=\int \frac{(x^2)^{n+1} -1}{x^2-1}$$

$$S=\frac{1}{x}\int \frac{(x^2)^{n+1} -1}{x^2-1}$$

Put $x=\frac {1}{4}$.

For $x\ne0,$ $$\dfrac{x^{2n}}{2n+1}=\dfrac1x\cdot\dfrac{x^{2n+1}}{2n+1}$$

Now for $-1\le x<1,$ (proof) $$\ln(1-x)=-\sum_{r=1}^\infty\dfrac{x^r}r$$

$\ln(1+x)-\ln(1-x)=?$

You can proof it in that way:

$${1\over 2}[1+{1 \over 3}({1 \over 4})^2+ {1 \over 5}({1 \over 4})^4+...] < {1\over 2}[1+({1 \over 4})^2+ ({1 \over 4})^4+...]$$ and this series also converges (part in brackets is a geometric series).

Edit

You can go starting with formula for inverse hyperbolic tangent function:

inverse hyperbolic tangent as similar series

which is:

$$\tanh^{-1}z = \sum_{n=1}^{\infty} {z^{2n-1} \over 2n-1}$$

And the answer is a very small modification of it:

$$4 \cdot \tanh^{-1}({1 \over 4}) = 4 \sum_{n=1}^{\infty} {({1 \over 4})^{2n-1} \over 2n-1} = 4 \cdot ({1 \over 1}({1 \over 4})^{1} + {1 \over 3}({1 \over 4})^{3} + ...) = 1 + {1 \over 3}({1 \over 4})^{2} + {1 \over 5}({1 \over 4})^{4} + ...$$