The Question

Does $\sum \frac{(n+4^n)}{n+6^n}$ converge or diverge?

Please note I have no knowledge of Alternating Series, Ratio and Root tests, Power Series, or Taylor and McLaurin Series.

My Work

The integral test didn't really appeal to me cause I wasn't sure how to take that integral. If that is the easiest approach and someone wants to help me with the integral then please show me the way.

By taking the simple Divergence test (the limit as n goes to infinity) I got a result of $0$ which unfortunately tells me nothing.

I can't really find a good value for my Comparison tests either. Can anyone help with any of these methods?


Hint: $\frac{n+4^n}{n+6^n} \leq\frac{n+4^n}{6^n} \leq 2(\frac{2}{3})^n$

The last is because $n < 4^n$. Can you do the rest?

  • $\begingroup$ And since $2\sum (\frac{2}{3})^n$ is a geometric series where $r < 1$ our original series must also be convergent? $\endgroup$ – Dunka Nov 11 '14 at 6:57
  • $\begingroup$ This is a good answer. I'm a little unclear as to why $n < 4^n$ shows the final inequality, could you show a few more steps on this please? $\endgroup$ – Dunka Nov 11 '14 at 7:08
  • 2
    $\begingroup$ $\frac{n+4^n}{6^n}\leq\frac{4^n+4^n}{6^n}=\frac{2\cdot4^n}{6^n}=2(4/6)^n$ $\endgroup$ – NickC Nov 11 '14 at 7:10

HINT: d'Alembert test (do you know it?) and $\frac23<1$.

Without d'Alembert test: $$ \frac{(n+4^n)}{n+6^n}\leq\frac{2\cdot4^n}{6^n}, $$ for large $n$ (in fact: always), and geometric sequence you should know.

  • $\begingroup$ I'm afraid I'm new to sequences and series, and do not know what the d'Alembert test is, thanks for the effort though :( $\endgroup$ – Dunka Nov 11 '14 at 6:48
  • $\begingroup$ How do you get that inequality? The solution is easy once I know that. $\endgroup$ – Dunka Nov 11 '14 at 7:10
  • 1
    $\begingroup$ @Dunka The background is to observe, that $n$ is meaningless in compare to $4^n$, hence we are trying to compare our series with some geometric series multiplied by a constant. So $n+4^n\leq4^n+4^n$ and obviously $n+6^n>6^n$, which gives even strong inequality. $\endgroup$ – Przemysław Scherwentke Nov 11 '14 at 7:15

Them limit will be similar to the limit of them largest growing terms. That would be $(4/6)^n$.


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