I have accrossed the following sum in my textbook $\lim\limits_{n\to \infty}\dfrac{\sum_{k=1}^{n} \left(\frac1k\right)}{\sum_{k=1}^{n}{\sin \left(\frac1k\right)}}=1$ , I have tried to evaluate nominator which it gives me $H_n$ n th Harmonic number , and in the denominator i have used this approach ${\sin \left(\frac1k\right)}\sim \frac1k $ for $k \to \infty $ but i can't get $1$. Any way ?
How can i prove this $\lim\limits_{n\to \infty}\dfrac{\sum_{k=1}^{n} \left(\frac1k\right)}{\sum_{k=1}^{n}{\sin \left(\frac1k\right)}}=1$
calculusharmonic-numbers
Related Solutions
To find the desired series we must first consider the following integral. $$\int _0^1\frac{\ln ^2\left(1-x\right)\operatorname{Li}_3\left(\frac{x}{x-1}\right)}{x}\:dx$$ To evaluate it one can make use of the following trilogarithm identity. $$\operatorname{Li}_3\left(\frac{x}{x-1}\right)=-\operatorname{Li}_3\left(x\right)-\operatorname{Li}_3\left(1-x\right)+\zeta \left(3\right)+\frac{1}{6}\ln ^3\left(1-x\right)$$ $$+\zeta \left(2\right)\ln \left(1-x\right)-\frac{1}{2}\ln \left(x\right)\ln ^2\left(1-x\right)$$ Using it on the previous integral yields: $$\int _0^1\frac{\ln ^2\left(1-x\right)\operatorname{Li}_3\left(\frac{x}{x-1}\right)}{x}\:dx=-\int _0^1\frac{\ln ^2\left(1-x\right)\operatorname{Li}_3\left(x\right)}{x}\:dx-\int _0^1\frac{\ln ^2\left(1-x\right)\operatorname{Li}_3\left(1-x\right)}{x}\:dx$$ $$+\zeta \left(3\right)\int _0^1\frac{\ln ^2\left(1-x\right)}{x}\:dx+\frac{1}{6}\int _0^1\frac{\ln ^5\left(1-x\right)}{x}\:dx+\zeta \left(2\right)\int _0^1\frac{\ln ^3\left(1-x\right)}{x}\:dx$$ $$-\frac{1}{2}\int _0^1\frac{\ln \left(x\right)\ln ^4\left(1-x\right)}{x}\:dx$$ $$=-\frac{81}{2}\zeta \left(6\right)+2\zeta ^2\left(3\right)+12\sum _{k=1}^{\infty }\frac{H_k}{k^5}-\sum _{k=1}^{\infty }\frac{H_k^2}{k^4}-\sum _{k=1}^{\infty }\frac{H_k^{\left(2\right)}}{k^4}-2\sum _{k=1}^{\infty }\frac{H_k^{\left(3\right)}}{k^3}$$ The series remaining can be calculated quite easily and a nice thing to know is that to evaluate them one does not have to cross paths with the series in the body of the OP.
Thus: $$\int _0^1\frac{\ln ^2\left(1-x\right)\operatorname{Li}_3\left(\frac{x}{x-1}\right)}{x}\:dx=-\frac{581}{24}\zeta \left(6\right)-4\zeta ^2\left(3\right)$$ Now what is left to do is to consider the following generating function. $$\sum _{k=1}^{\infty }\frac{x^{k-1}}{k}\left(H_k^2+H_k^{\left(2\right)}\right)=-2\frac{\operatorname{Li}_3\left(\frac{x}{x-1}\right)}{x}$$ Which can be found along other generating functions in the book (Almost) Impossible Integrals, Sums, and Series, page $\#285$.
Using it on the previously found integral means that, $$\int _0^1\frac{\ln ^2\left(1-x\right)\operatorname{Li}_3\left(\frac{x}{x-1}\right)}{x}\:dx=-\frac{1}{2}\sum _{k=1}^{\infty }\left(\frac{H_k^2+H_k^{\left(2\right)}}{k}\right)^2$$ Thus: $$\sum _{k=1}^{\infty }\left(\frac{H_k^2+H_k^{\left(2\right)}}{k}\right)^2=\frac{581}{12}\zeta \left(6\right)+8\zeta ^2\left(3\right)$$
Using the Euler–Boole summation formula, $$ \sum\limits_{n = 1}^\infty {\frac{{( - 1)^n }}{{H_n }}} \sim \sum\limits_{n = 1}^{N - 1} {\frac{{( - 1)^n }}{{H_n }}} + ( - 1)^N \frac{1}{2}\sum\limits_{k = 0}^\infty {\frac{{E_{2k} }}{{4^k (2k)!}}\left[ {\frac{{{\rm d}^{2k} }}{{{\rm d}x^{2k} }}\frac{1}{{\psi (x + 1/2) + \gamma }}} \right]_{x = N} } $$ as $N\to +\infty$. Here, $E_k$ denotes the Euler numbers, $\gamma$ is the Euler–Mascheroni constant, and $\psi(x)$ is the digamma function. Taking $N=100$ and keeping the first $5$ terms of the asymptotic expansion, I obtained $-0.626332482737912\ldots$ for the value of your sum.
Best Answer
The sequences $$ a_n =\sum_{k=1}^{n}{\sin \frac1k} \,, \, b_n = \sum_{k=1}^{n} \frac1k $$ satisfy the conditions of the Stolz–Cesàro theorem: $(b_n)$ is strictly increasing and divergent, and $$ \lim_{n \to \infty} \frac{a_n - a_{n-1}}{b_n - b_{n-1}} = \lim_{n \to \infty} \frac{\sin \frac1n}{\frac1n} = 1 $$ exists. The theorem then states that $$ \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{a_n - a_{n-1}}{b_n - b_{n-1}} = 1 \, . $$