Convergence and Divergence of Series — Question 10

*Question 10 Determine whether $\displaystyle\sum_{n=1}^{\infty}\frac1{n^2+\sin n}$ converges or diverges.

1. Show that the denominator is positive for every $n\ge1$.

2. Use $-1\le\sin n\le1$ to produce a simple upper comparison for the tail.

3. Confirm the classification by limit comparison with $1/n^2$.

*Question 10 – Solution

For $n=1$, $1+\sin1>0$. For $n\ge2$, use $\sin n\ge-1$ to obtain $$n^2+\sin n\ge n^2-1>0,$$ so every term is defined and positive.

For $n\ge2$, $$n^2-1\ge\frac34n^2,$$ because $n^2\ge4$. Hence $$0<\frac1{n^2+\sin n}\le\frac1{n^2-1}\le\frac4{3n^2}.$$ The comparison series is a constant multiple of the convergent $p$-series $\sum1/n^2$. The Direct Comparison Test therefore proves convergence of the tail, and adding the finite first term cannot change that conclusion.

Limit comparison gives the sharper asymptotic statement $$\lim_{n\to\infty}\frac{1/(n^2+\sin n)}{1/n^2} =\lim_{n\to\infty}\frac{n^2}{n^2+\sin n}=1,$$ because dividing numerator and denominator by $n^2$ produces $1/(1+\sin n/n^2)$, and $|\sin n|/n^2\le1/n^2\to0$. The ratio limit $1$ confirms that the oscillation in the denominator is asymptotically negligible.