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determine the convergence of $$ \sum_{n=1}^{\infty} \frac{n !}{n !+3} $$

I tried using the ratio test and also for n! , I use Stirling approximation.Still I got stuck.

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    A necessary condition for $\sum_{n\geq 1}a_n$ to converge is $\lim_{n\to +\infty}a_n=0$, which is clearly not fulfilled in your case. – Jack D'Aurizio Jun 15 '20 at 10:39

3 Answers3

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A series $\sum a_n$ cannot converge unless $a_n \to 0$. Here $a_n \to 1$ so it is not convergent.

Kavi Rama Murthy
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Since $\lim_{n\to\infty}\frac{n!}{n!+3}=1\ne0$, your series diverges.

José Carlos Santos
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  • Sorry your answer is not cleared to me –  Jun 15 '20 at 10:38
  • I am just using the fact that, if a series $\sum_{n=1}^\infty a_n$ converges, then $\lim_{n\to\infty}a_n=0$. In your case, that limit is $1$, not $0$. Is it clear now? – José Carlos Santos Jun 15 '20 at 10:40
  • yes! thanks .It is now clear to me –  Jun 15 '20 at 10:43
  • So, when to use ratio test / root test etc... actually I am getting confused to deal with the inspection to study the nature of series! –  Jun 15 '20 at 10:46
  • I am sorry, but there is no test to decide which test you should use. However, if $a_n$ is something raised to the power $n$, I suggest that you use the root test. And if $a_n$ is a product, I suggest that you use the ratio test. But the most basic of all tests os to determine whether or not $\lim_{n\to\infty}a_n=0$. – José Carlos Santos Jun 15 '20 at 10:50
  • clear! thanks a lot Sir. –  Jun 15 '20 at 10:52
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Upper bound (take $n!+3 \geq n!$): $a_n \leq 1$, so $\sum_{n}a_n \to_n \infty$

Lower bound (take $n! + 3 \leq 2 n!): a_n \geq \frac{1}{2}$, so $\sum_n a_n \to_n \infty$

So the sum diverges. In fact only the lower bound would do.

Alex
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