Question

question 12 of 13 > determine convergence or divergence of $sum_{n = 1}^{infty}\frac{1}{n^{2}+sin(n)}$ using any method covered so far. the series diverges. the series converges.

Explanation:

Step1: Use the comparison test

We know that \(- 1\leqslant\sin(n)\leqslant1\) for all \(n\). Then \(n^{2}+\sin(n)\leqslant n^{2} + 1\). So \(\frac{1}{n^{2}+\sin(n)}\geqslant\frac{1}{n^{2}+1}\). Also, for large \(n\), we compare with the \(p -\)series. We know that \(\sum_{n = 1}^{\infty}\frac{1}{n^{2}}\) is a convergent \(p -\)series with \(p = 2>1\). And \(\frac{1}{n^{2}+\sin(n)}\leqslant\frac{1}{n^{2}- 1}\) for \(n\geqslant2\). We use the limit - comparison test. Let \(a_{n}=\frac{1}{n^{2}+\sin(n)}\) and \(b_{n}=\frac{1}{n^{2}}\).

Step2: Calculate the limit

\(\lim_{n
ightarrow\infty}\frac{a_{n}}{b_{n}}=\lim_{n
ightarrow\infty}\frac{\frac{1}{n^{2}+\sin(n)}}{\frac{1}{n^{2}}}=\lim_{n
ightarrow\infty}\frac{n^{2}}{n^{2}+\sin(n)}=\lim_{n
ightarrow\infty}\frac{1}{1+\frac{\sin(n)}{n^{2}}}\)
Since \(\lim_{n
ightarrow\infty}\frac{\sin(n)}{n^{2}} = 0\) (because \(|\sin(n)|\leqslant1\) and \(\lim_{n
ightarrow\infty}\frac{1}{n^{2}}=0\)), then \(\lim_{n
ightarrow\infty}\frac{a_{n}}{b_{n}} = 1\).

Step3: Apply the limit - comparison test result

Since \(\sum_{n = 1}^{\infty}b_{n}=\sum_{n = 1}^{\infty}\frac{1}{n^{2}}\) is a convergent \(p -\)series (\(p = 2>1\)) and \(\lim_{n
ightarrow\infty}\frac{a_{n}}{b_{n}}=1>0\), by the limit - comparison test, \(\sum_{n = 1}^{\infty}\frac{1}{n^{2}+\sin(n)}\) converges.

Answer:

The series converges.