Given the series:

$\sum_{n=1}^\infty \left(\frac {(4n + 7)}{(n + 3)}\right)^n $

How to check it converges?

  • $\begingroup$ Just use root test. And then the inside limit is $4$ and so it diverges. $\endgroup$ – Fundamental Feb 25 '16 at 3:35
  • $\begingroup$ the general term diverges $\endgroup$ – reuns Feb 25 '16 at 3:53
  • $\begingroup$ @Fundamental seems root test fits this question. $\endgroup$ – vivek Feb 25 '16 at 4:01

If $\sum_{n=1}^\infty a_n$ converges, then that implies that $a_n\to 0$ as $n\to\infty$.

Via contrapositive, if $a_n\not\to 0$ then $\sum_{n=1}^\infty a_n$ does not converge.

In your case, your summands are of the form $(\frac{4n+7}{n+3})^n$ which is on the order of $O(4^n)$ and does not approach zero as $n$ grows large. You are adding infinitely many large things together, so it follows that the sum will diverge to infinity.


Using the root test

$$ \lim_{n\to\infty} \left(\left(\frac {4n+7}{n+3} \right)^n\right)^{1/n} $$

$$ \lim_{n\to\infty} \left(\frac {4n+7}{n+3} \right) $$

$$ \lim_{n\to\infty} \left(\frac {4 + 7/n}{1 + 3/n} \right) $$

Applying the limit:

$$ \lim_{n\to\infty} (7/n) \to 0 \text{ and} \lim_{n\to\infty} (3/n) \to 0 $$

$$ \left(4/1\right) = 4 > 1 $$ hence, the series diverges.


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