Help me with this Series Problem (Comparison Test)?

It diverges. The numerator is exponential but the denominator is only polynomial of degree 8, which means that the terms themselves will diverge too. In other words

lim n→∞ (4+3ⁿ)/(1+n²)⁴
= lim n→∞ (ln 3)⁸3ⁿ/8! (after 8 applications of l'Hôpital's rule)
= ∞.

Apply the ratio test to get

L = lim (n→∞) 3ⁿ⁺¹/3ⁿ * (n/(n+1))⁸ = 3*1 = 3

Because L>1, by the test the series is divergent

You could have done the same with the original series.


If you want to avoid the test, another way is to see that

3ⁿ⁺¹/(n+1)⁸ > 3ⁿ/n⁸ if 3 > (1+1/n)⁸

⟹ ⁸√3−1 > 1/n ⟹ n > 1/(⁸√3−1) ≈ 6.79

So for n>6 the terms increase so the series must diverge

　
For large n, exponential functions grow much faster than power function.

To show that 3^n/n^8 diverges, take limit.
You will get ∞/∞, so you can apply L'Hopital's Rule.
You'll have to do this several times.

Each time, numerator becomes 3^n * (ln 3) * (ln 3) ...
but denominator gets smaller powers for n, until eventually you get:
3^n * (ln 3)^8 / 8!, which clearly has limit = ∞

So 3^n/n^8 diverges