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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 ?

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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 $$ \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{a_n - a_{n-1}}{b_n - b_{n-1}} = 1 \, . $$

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  • $\begingroup$ It is interesting that both versions are possible, but the one with the sine function in the denominator is a little bit more difficult because die divergence is not common knowledge like for the harmonic sum. Nice answer btw :). $\endgroup$ – MachineLearner Mar 14 at 21:26
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Hint: $\sin(1/k) = 1/k + O(1/k^3)$, and $\sum_{k=1}^\infty 1/k^3$ converges.

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You already reacived good and simple answer to the question. So, let us try to find the asymptotics for $$y_n=\dfrac{\sum_{k=1}^{n} \left(\frac1k\right)}{\sum_{k=1}^{n}{\sin \left(\frac1k\right)}}$$

By Taylor, we have $$\sin \left(\frac1k\right)=\sum_{m=1}^\infty \frac{(-1) ^m}{(2m+1)!}k^{-(2m+1)}=\frac{1}{k}-\frac{1}{6 k^3}+\frac{1}{120 k^5}-\frac{1}{5040 k^7}+O\left(\frac{1}{k^9}\right)$$ Then $$\sum_{k=1}^{n}{\sin \left(\frac1k\right)}=H_n-\frac{H_n^{(3)}}{6}+\frac{H_n^{(5)}}{120}-\frac{H_n^{(7)}}{5040}+\cdots$$ where appear generalized harmonic numbers. So $$y_n=\frac{H_n}{H_n-\frac{H_n^{(3)}}{6}+\frac{H_n^{(5)}}{120}-\frac{H_n^{(7)}}{5040}+\cdots}$$ Using their expansion up to $O\left(\frac{1}{n^6}\right)$ we then have $$y_n=\frac{\gamma +\log \left({n}\right)+\frac{1}{2 n}-\frac{1}{12 n^2}+\frac{1}{120 n^4}+O\left(\frac{1}{n^6}\right) } {\left(\gamma +\frac{\psi ^{(2)}(1)}{12}-\frac{\psi ^{(4)}(1)}{2880}+\frac{\psi ^{(6)}(1)}{3628800}\right)+\log(n)+\frac{1}{2 n}-\frac{1}{12 n^3}+\frac{23}{480 n^4}+\frac{1}{240 n^5}+O\left(\frac{1}{n^6}\right) }$$ Computing for a few values of $n$ $$\left( \begin{array}{ccc} n & \text{approximation} & \text{exact} \\ 1 & 1.1734669 & 1.1883951 \\ 2 & 1.1353960 & 1.1355924 \\ 3 & 1.1123848 & 1.1123980 \\ 4 & 1.0990964 & 1.0990971 \\ 5 & 1.0903323 & 1.0903314 \\ 6 & 1.0840428 & 1.0840417 \\ 7 & 1.0792645 & 1.0792633 \\ 8 & 1.0754827 & 1.0754816 \\ 9 & 1.0723966 & 1.0723955 \\ 10 & 1.0698175 & 1.0698165 \end{array} \right)$$

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