If \(\lim_{n\to\infty}a_n\neq0\) or does not exist, then the infinite series \(\sum_{n=1}^\infty a_n\) is divergent.
If \(\mathrm\lim_{n\to\infty}a_n=0\mathrm{~,}\) then nothing can be said about the convergence of the infinite series.
\(\sum_{n=1}^\infty a_n\), i.e., the series could converge or diverge.
Example. Determine whether each infinite series converges or diverges.
\(\begin{aligned}(\mathrm{a})&&\sum_{n=1}^\infty\arctan(n)&&(\mathrm{b})&&\sum_{n=1}^\infty e^n\\\\(\mathrm{c})&&\sum_{n=1}^\infty e^{-n}&&(\mathrm{d})&&\sum_{n=1}^\infty\frac{n+1}n\end{aligned}\)
Solution.
\(\begin{aligned}(\mathrm{a})&\quad\mathrm{Let~}a_n=\arctan(n).\\&\lim_{x\to\infty}a_n=\lim_{x\to\infty}\arctan(n)=\frac\pi2\neq0\end{aligned}\)
By the Divergence Test, the series diverges.
\(\begin{aligned}\mathrm{(b)}\quad&\quad\mathrm{Let~}a_n=e^n.\\&\lim_{x\to\infty}a_n=\lim_{x\to\infty}e^n=\infty\neq0\end{aligned}\)
By the Divergence Test, the series diverges.
\(\mathrm{(c)\quad~Let~}a_n=e^{-n}.\)
\(\lim_{x\to\infty}a_n=\lim_{x\to\infty}e^{-n}=0\)
By the Divergence Test, no conclusion can be made about the convergence of the series. Try another series test.
\(\begin{aligned}(\mathrm{d})&\quad\mathrm{Let~}a_n=\frac{n+1}n.\\&\lim_{x\to\infty}a_n=\lim_{x\to\infty}\frac{n+1}n=\lim_{x\to\infty}\frac{1+1/n}1=1\neq0\end{aligned}\)
By the Divergence Test, the series diverges.