A calculus problem by Christian Daang

Let the required sum be $S$ . Consider the following sum:

$\begin{aligned} Z & = \frac 11 + \frac i2 - \frac 13 - \frac i4 + \frac 15 + \frac i6 - \frac 17 - \frac i8 + \frac 19 + \frac i{10} - \frac 1{11} - \frac i{12} + ... \\ \implies S & = \Re(Z) + \Im(Z) \end{aligned}$

Note that:

$\begin{aligned} Z & = \frac 11 + \frac i2 - \frac 13 - \frac i4 + ... \\ & = \frac {i^0}1 + \frac {i^1}2 + \frac {i^2}3 + \frac {i^3}4 + ... \\ & = \frac 1i \left( \frac {i^1}1 + \frac {i^2}2 + \frac {i^3}3 + \frac {i^4}4 + ... \right) & \small \color{#3D99F6} \text{Note that }\ln (1-x) = - \sum_{k=1}^\infty \frac {x^k}k \\ & = i \ln (1-i) \\ & = i \ln \left(\sqrt 2 e^{-\frac \pi 4i} \right) \\ & = i \left(\ln \sqrt 2 - \frac \pi 4 i \right) \\ & =\frac \pi 4 + i \ln \sqrt 2 \end{aligned}$

$\implies S = \Re(Z) + \Im(Z) = \frac \pi 4 + \ln \sqrt 2 \approx \boxed{1.132}$

the expression above is equal to:

$\begin{aligned} \sum_{n = 0}^{\infty} \left( \cfrac{1}{4n+1} + \cfrac{1}{4n+2} - \cfrac{1}{4n+3} - \cfrac{1}{4n+4}\right) \\ & = \sum_{n = 0}^{\infty} \left( \int_0^1 (1+x-x^2-x^3)\left(x^{4n}\right) \ dx \right) \end{aligned}$

If we will try to expand it, it will be like:

$\begin{aligned} \left( \int_0^1 (1+x-x^2-x^3)\left(x^{0}\right) \ dx \right) + \left( \int_0^1 (1+x-x^2-x^3)\left(x^{4}\right) \ dx \right) + \left( \int_0^1 (1+x-x^2-x^3)\left(x^{8}\right) \ dx \right) + \left( \int_0^1 (1+x-x^2-x^3)\left(x^{12}\right) \ dx \right) + \left( \int_0^1 (1+x-x^2-x^3)\left(x^{16}\right) \ dx \right) + \cdots \\ & = \left( \int_0^1 (1+x-x^2-x^3)\left(x^{0} + x^4 + x^8 + x^{12} + x^{16} + \cdots \right) \ dx \right) \\ & = \left( \int_0^1 (1+x-x^2-x^3)\left(\sum_{n = 0}^{\infty} x^{4n}\right) \ dx \right) \\ & = \int_0^1 \cfrac{(1+x-x^2-x^3)}{1 - x^4} \ dx \\ & = \int_0^1 \cfrac{(1+x)(1-x^2)}{(1-x^2)(1+x^2)} \ dx \\ & = \int_0^1 \cfrac{(1+x)}{(1+x^2)} \ dx \\ & = \left. \cfrac{\ln (x^2 + 1)}{2} + \arctan (x) \right |_0^1 \\ & = \boxed{ \cfrac{\ln (2)}{2} + \cfrac{\pi}{4} \approx 1.13197175367742096432 } \end{aligned}$

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Or

$\begin{aligned} \sum_{n = 1} ^{\infty} \left(\cfrac{1}{(4n - 3)} + \cfrac{1}{(4n - 2)} - \cfrac{1}{(4n - 1)} - \cfrac{1}{(4n)}\right) \\ &= \sum_{n = 1}^{\infty} \left(\int_0^1 \left(x^{4n-4} + x^{4n-3} - x^{4n-2} - x^{4n-1}\right) \ dx \right) \end{aligned}$

If we will try to expand it, it will be like:

$\begin{aligned} \left(\int_0^1 \left(x^{0} + x^{1} - x^{2} - x^{3}\right) \ dx \right) + \left(\int_0^1 \left(x^{4} + x^{5} - x^{6} - x^{7}\right) \ dx \right) + \left(\int_0^1 \left(x^{8} + x^{9} - x^{10} - x^{11}\right) \ dx \right) + \cdots \\ & = \int_0^1 \left( \big(x^{0} + x^{1} - x^{2} - x^{3}\big) + \big(x^{4} + x^{5} - x^{6} - x^{7}\big) + \big(x^{8} + x^{9} - x^{10} - x^{11}\big) + \cdots \right) \ dx \\ & = \int_0^1 \left( \big(x^{0} + x^{4} + x^{8} + x^{12} + \cdots\big) + \big(x^{1} + x^{5} + x^{9} + x^{13} + \cdots\big) - \big(x^{2} + x^{6} + x^{10} + x^{14} + \cdots \big) - \big(x^{3} + x^{7} + x^{11} + x^{15} + \cdots \big) \right) \ dx \\ & = \int_0^1 \left(\sum_{n = 1}^{\infty} \left(x^{4n-4} + x^{4n-3} - x^{4n-2} - x^{4n-1} \right) \right) \ dx \\ &= \int_0^1 \left( \cfrac{1}{(1-x^4)} + \cfrac{x}{(1-x^4)} - \cfrac{x^2}{(1-x^4)} - \cfrac{x^3}{(1-x^4)} \right) \ dx \\ &= \int_0^1 \cfrac{1+x-x^2 - x^3}{(1-x^4)} \ dx \end{aligned}$

And the result is just the same as the result on the above solution.

From well known Log series

$\ln (1+x)=x-x^2/2+x^3/3-x^4/4+.....+ \infty$

Put $x=i$ and put $x=-i$ where $i=\sqrt{-1}$

Adding these two we get $\ln(1+i) + \ln(1-i) = 2(1/2 - 1/4+1/6-1/8 + ..... + \infty)$

Now subtract $\ln(1+i) - \ln(1-i)=2i(1-1/3+1/5-1/7+..... \infty)$

Value of the first series $\ln ((1+i)(1-i))/2=\ln 2/2$

Value of the second series $\ln(\dfrac{1+i}{1-i})/2i=ln(i)/2i=\dfrac{i \pi/2}{2i}=\dfrac{\pi}{4}$

So, total value of the series $=\dfrac{1}{2} \ln(2) + \dfrac{\pi}{4}=1.131$

Note All calculations are not shown.User can easily check it by themselves.