Hey there.. i've done the ration test for the series at the attachment..
i've got the answer 1/7 where the series is convergent..
can someone confirm my answer..?
Pretty straight forward, isn't it? $\displaystyle a_n= \frac{n^3}{7^n}$ so $\displaystyle a_{n+1}= \frac{(n+1)^3}{7^{n+1}}$ and the ratio is $\displaystyle \frac{a_{n+1}}{a_n}= \frac{(n+1)^3}{7^{n+1}}\frac{7^n}{n^3}$$\displaystyle = \frac{(n+1)^3}{n^3}\frac{7^n}{7^{n+1}}= \frac{n^3+ 3n^2+ 3n+ 1}{n^3}\frac{1}{7}= \left(1+ \frac{3}{n}+ \frac{3}{n^2}+ \frac{1}{n^3}\right)\left(\frac{1}{7}\right)$.
As n goes to infinity, those fractions go to 0 so the limit $\displaystyle (1)(\frac{1}{7})< 1$. Yes, the series converges.