$$\sum_{n=1}^{\infty}\frac{(n+2)!}{(3n-1)}$$ I know this series does not converge. Can someone show me how to prove that? Should i use criteria of Dalamber or any other criteria?
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I've edited your question to fix the MathJax; please verify that it says what you intended. – May 09 '14 at 18:43
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1http://en.wikipedia.org/wiki/Ratio_test – lab bhattacharjee May 09 '14 at 18:43
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1You could show the terms of the series do not approach $0$ as $n$ tends to infinity. (Note $(n+2)!>n^2$.) – David Mitra May 09 '14 at 18:44
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1Wow does that ever not converge! Explosive blow up. – kleineg May 09 '14 at 18:54
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The part of the series itself diverges. Use integral test to check it out. Forget about the whole series. $$\sum_{n=1}^{\infty}\frac{1}{(3n-1)} \le \sum_{n=1}^{\infty}\frac{(n+2)!}{(3n-1)}$$ also note that the first necessity for convergence of series is that it's limit $n\to\infty$ must be zero. i.e, $$\lim_{n\to\infty}a_n = 0$$
S L
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The Ratio Test is usually the best way to approach factorials in series, however with this one you can just use the Test for Divergence by showing that the terms of the series do not approach $0$. In fact, the terms should all be greater than $1$, because the numerator is greater than the denominator. That's pretty straightforward with induction.
In general, your method should be:
Verify that the terms approach $0$ and that the absolute value of the terms are decreasing, either by showing that $a_{n+1}<a_n$ or by showing that the derivative is negative. If not, then use the Test for Divergence.
If the series is alternating, use the Alternating Series Test. If the terms approach $0$ and their absolute value is decreasing, it is at least conditionally convergent.
Check if the series is a geometric series, a telescoping series (usually by partial fraction decomposition), or a $p$-series (of the form $\frac1{n^p}$).
Geometric series converge when the ratio $|r|<1$. $p$-series converge when $p>1$.
If the series looks like a rational function, use the Comparison Tests against a $p$-series. Pull the leading terms out and cancel them down to find the appropriate exponent to use.
If the Comparison Test fails in the last case, because it's greater when it should be less than or vice versa, then use the Limit Comparison Test (or just jump straight here and skip the last one).
If the series has $\sin(n)$ or $\cos(n)$ in it, use the Comparison Test, replacing the trig function with $\pm1$.
If the series has factorials in it, use the Ratio Test (also apparently known as the criteria of Dalamber).
If the series has an exponent around a nasty term, use the Root Test.
If all else fails, or if you think the integral is easy enough, use the Integral Test. That's how most of these tests are proved anyway.
Zook
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If you prove that $(n+2)!$ grows faster than $(3n-1)$, the general term will grow, and the summatory will not converge. So, you can prove this inequality (by induction):
\begin{equation} ((n+1)-2)!- (n-2)! > (3(n+1)-1) - (3n-1) \end{equation}
This proof method is easy because $a_{n+1} > a_n$.
