Calc6.25

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Discuss the convergence or divergence of the series \sum _{{n=1}}^{{\infty }}{\frac  {n+1}{n\times 3^{{n+1}}}}.

Let us compare this series with the geometric series \sum _{{n=1}}^{{\infty }}\left({\frac  {1}{3}}\right)^{n}.

If r < 1, then the series converges. If r > 1, then the series diverges. If r = 1, the ratio test is inconclusive, and the series may converge or diverge. Since in our case, r is equal to 1/3, the geometric series must converge.

\lim _{{n\rightarrow \infty }}{\frac  {{\frac  {1}{3^{n}}}}{{\frac  {n+1}{n3^{{n+1}}}}}}=\lim _{{n\rightarrow \infty }}{\frac  {3n}{n+1}}=3\,

Thus, since this limit is finite and positive, both series converge or diverge. Since the geometric series converges, then our series must also converge.


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