A calculus problem by Sarth Vitekar

The $N$ th partial sum can be written as $\begin{aligned} S_N &= \sum_{n=1}^N \frac{1}{n} - \sum_{n=1}^{\lfloor N/3\rfloor} \frac{3}{3n} = \sum_{n=1}^N \frac{1}{n} - \sum_{n=1}^{\lfloor N/3\rfloor} \frac{1}{n} \\ &= \left(\sum_{n=1}^N \frac{1}{n} - \ln(N)\right) - \left(\sum_{n=1}^{\lfloor N/3\rfloor} \frac{1}{n} - \ln\lfloor{N/3}\rfloor\right) + \left(\ln(N) - \ln\lfloor{N/3}\rfloor\right) \\ &= \left(\sum_{n=1}^N \frac{1}{n} - \ln(N)\right) - \left(\sum_{n=1}^{\lfloor N/3\rfloor} \frac{1}{n} - \ln\lfloor{N/3}\rfloor\right) + \ln\left(\frac{N}{\lfloor{N/3}\rfloor}\right) \end{aligned}$

Letting $N\rightarrow\infty$ , and writing $\gamma$ for the Euler-Mascheroni constant, we have $\lim_{N\rightarrow\infty} S_N = \gamma - \gamma + \ln(3) = \ln(3) \approx \boxed{1.09861229}$

$\begin{aligned} S & = \frac 11 + \frac 12 - \frac 23 + \frac 14 + \frac 15 - \frac 26 + \frac 17 + \frac 18 - \frac 29 + ... \\ & = \frac {-2 \left(-\frac 12\right)}1+ \frac {-2 \left(-\frac 12\right)}2 - \frac {2(1)}3 + \frac {-2 \left(-\frac 12\right)}4 + \frac {-2 \left(-\frac 12\right)}5 - \frac {2(1)}6 + ... \\ & = - \frac {2\cos \frac {2\pi}3}1 - \frac {2\cos \frac {4\pi}3}2 - \frac {2\cos \frac {6\pi}3}3 - \frac {2\cos \frac {8\pi}3}4 - \frac {2\cos \frac {10\pi}3}5 - \frac {2\cos \frac {12\pi}3}5 - ... \\ & = \sum_{n=1}^\infty - 2\frac {\cos \frac {2n\pi}3}n \\ & = \sum_{n=1}^\infty \Re \left \{ - 2\frac {e^{\frac {2n\pi}3i}}n \right \} \\ & = - 2 \Re \left \{ \sum_{n=1}^\infty \frac {e^{\frac {2n\pi}3i}}n \right \} \\ & = 2 \Re \left \{\ln \left(1-e^{\frac {2n\pi}3i} \right) \right \} \\ & = 2 \Re \left \{\ln \left(1+\frac 12 - i\frac {\sqrt 3}2 \right) \right \} \\ & = 2 \Re \left \{\ln \left(\frac 32 - i\frac {\sqrt 3}2 \right) \right \} \\ & = 2 \Re \left \{\ln \left(\sqrt 3 e^{-i \tan^{-1}\frac 1{\sqrt 3}} \right) \right \} \\ & = 2 \Re \left \{\ln \sqrt 3 -i \tan^{-1}\frac 1{\sqrt 3} \right \} \\ & = 2 \ln \sqrt 3 \approx \boxed{1.0986} \end{aligned}$

$\displaystyle S = \frac{1}{1} + \frac{1}{2} - \frac{2}{3} + \frac{1}{4} + \frac{1}{5} - \frac{2}{6} + \cdots$

$\displaystyle S = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots - 3\left(\frac{1}{3} + \frac{1}{6} + \frac{1}{9} + \cdots \right)$

If we consider $\frac{x}{1} + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \cdots = \ln\left(\frac{1}{1-x}\right)$

$\displaystyle \ln\left(\frac{1}{1-\omega}\right) = \frac{\omega}{1} + \frac{\omega^2}{2} + \frac{\omega^3}{3} + \cdots$

$\displaystyle \ln\left(\frac{1}{1-\omega^2}\right) = \frac{\omega^2}{1} + \frac{\omega^4}{2} + \frac{\omega^6}{3} + \cdots$

$\displaystyle \lim_{x\rightarrow 1} \ln\left(\frac{1}{1-x}\right) = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \cdots$

$\displaystyle \implies \lim_{x\rightarrow 1} \ln\left(\frac{1}{1-x}\right) + \ln\left(\frac{1}{1-\omega^2}\right) + \ln\left(\frac{1}{1-\omega}\right) = 3\left(\frac{1}{3} + \frac{1}{6} + \frac{1}{9} + \cdots \right)$

Therefore,

$\displaystyle S = \lim_{x\rightarrow 1} \ln\left(\frac{1}{1-x}\right) - \lim_{x\rightarrow 1} \ln\left(\frac{1}{1-x}\right) - \ln\left(\frac{1}{1-\omega^2}\right) - \ln\left(\frac{1}{1-\omega}\right)$

$\displaystyle S = \log(1 - \omega - \omega^2 + 1) = \log(3)$

the expression above is equal to:

$\sum_{n = 0}^{\infty} \left( \cfrac{1}{3n+1} + \cfrac{1}{3n+2} - \cfrac{2}{3n+3} \right) \\ = \sum_{n = 0}^{\infty} \left( \int_0^1 (1+x-2x^2)\left(x^{3n}\right) \ dx \right) \\ = \int_0^1 \cfrac{(1+x-2x^2)}{1 - x^3} \ dx \\ = \int_0^1 \cfrac{2x+1}{x^2 + x + 1} \ dx \\ = \left. \ln (x^2 + x + 1) \right |_0^1 \\ = \boxed{ \ln (3) }$