0
$\begingroup$

I am asked to show whether the series

$$\sum_{i=1}^{\infty}\frac{n+7^n}{n+5^n}$$

is convergent or divergent. I tried using Comparison Test but I was unsure of what to compare it to since there is addition in both the numerator and denominator. I also tried factoring out n from both parts of the fraction but this was not helpful either. I could use Ratio Test but we did not learn it yet. At this point I am stuck and have no idea what other methods to try.

$\endgroup$
  • 5
    $\begingroup$ Hint: for convergent series, the terms must go to $0$. $\endgroup$ – lulu Nov 5 '19 at 21:19
1
$\begingroup$

Since$$\lim_{n\to\infty}\frac{n+7^n}{n+5^n}=\lim_{n\to\infty}\left(\frac75\right)^n\frac{n7^{-n}+1}{n5^{-n}+1}=\infty,$$your series diverges.

| cite | improve this answer | |
$\endgroup$
-1
$\begingroup$

Note that

$$\frac{n+7^n}{n+5^n}\ge \frac{7^n}{2\cdot5^n}=\frac12\cdot \left(\frac 7 5\right)^n \to \infty$$

therefore the series can't converge indeed for any $\sum_1^\infty a_n =L$ we have necessarly that

$$S_{N+1}-S_N=\sum_1^{N+1} a_n-\sum_1^N a_n=a_{N+1} \to L-L=0$$

| cite | improve this answer | |
$\endgroup$
-1
$\begingroup$

Apply limit comparison with $$\sum _1^\infty (7/5)^n$$ which is a geometric series with $r=7/5$.

Note that $$\lim _{n\to \infty} \frac {(\frac {n+7^n}{n+5^n})}{ (\frac {7}{5})^n} =1$$

Thus the series $$\sum_{i=1}^{\infty}\frac{n+7^n}{n+5^n}$$ diverges as well.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.