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For positive integers $n$, the summand is $a_n=\sin (\pi \sin (1/n))/n$. Since $0<\sin (1/n)<1$, these terms are positive. Apply the limit comparison test with the convergent p-series $\sum n^{-2}$:Each sine ratio tends to one. The limit comparison test says that two positive-term series whose term ratio tends to a finite positive number are either both convergent or both divergent. ThereforeThere is also a direct comparison test for series: $0\le \sin (\pi \sin (1/n))\le \pi \sin (1/n)\le \pi /n$, giving $0\le a_n\le \pi /n^2$.

Suppose real coefficients satisfy $c\le a_n\le C(1+n{)}^m$ for all sufficiently large $n$, where $c,C>0$ and $m\ge 0$. Then the power series $\sum a_n{z}^n$ has radius of convergence one. For $|z|<1$, the comparison test for series and the ratio test dominate it by a polynomial times a geometric sequence. For $|z|\ge 1$, the terms have modulus at least $c$ eventually and violate the term test for divergence. Finite changes to the coefficients do not affect the conclusion.

Suppose $a_n>0$ and, for all sufficiently large $n$, $a_{n+1}/a_n\le 1-c/n$ with fixed $c>0$. Choose an initial index $N>c$. Since $\log (1-x)\le -x$ for $0<x<1$, summing logarithmic ratios gives $\log (a_n/a_N)\le -c{\sum }_{k=N}^{n-1}1/k\le -c\log (n/N)$. Thus $a_n\le a_N{N}^c{n}^{-c}$. For $c>1$, the comparison test for series with a P-series proves convergence. This supplies a useful refinement when a limiting ratio of one makes the ordinary ratio test inconclusive.

For positive sequences $(a_n),(b_n)$, suppose $a_{n+1}/a_n\le b_{n+1}/b_n$ eventually. Then $a_n/b_n$ is eventually nonincreasing, so $a_n\le C{b}_n$ beyond a fixed index. The comparison test for series proves convergence of $\sum a_n$ whenever $\sum b_n$ converges. Reversing the ratio inequality gives an eventual lower bound $a_n\ge c{b}_n$ and hence divergence when the positive comparison series diverges. The initial finite set of terms does not affect either conclusion.

For $p>0$ and $n\ge 2$, integrating the decreasing derivative magnitude of $x^{-p}$ gives $(n-1{)}^{-p}-n^{-p}=p{\int }_{n-1}^n{x}^{-p-1}dx>p{n}^{-p-1}$. Put $p=k-1$ and sum this telescoping series to obtain the bound in the title, strict for $N>1$. The partial sums are increasing and bounded, proving convergence of the p-series when $k>1$. This gives a direct comparison test for series proof and a finite upper bound without evaluating the sum.

Let a power series ${\sum }_{n\ge 0}{a}_n{z}^n$ have finite radius of convergence $R$. If its terms were bounded in complex modulus at a point $z$ with $|z|>R$, say $|a_n{z}^n|\le M$, choose $w$ with $R<|w|<|z|$. ThenThe comparison test for series with a convergent geometric series would give absolute convergence at $w$, contradicting its position outside the convergence disc. Hence the terms are unbounded at every such $z$. The argument also works when $R=0$ and gives a stronger obstruction than the usual term test for divergence.