Question

summation from n=2 to infinity (n^.5 +4)/n^2. Prove it's convergent by comparison.

Answer 1

for large values of n, this is pretty much sqrt(n)/n^2, or 1/n^(3/2)

Answer 2

\[ \frac{n^.5 +4}{n^2}<\frac{1}{n}^{(b)}\]
\[ n^b<\frac{n^2}{n^.5 +4}\]
\[ ln(n^b)<ln(\frac{n^2}{n^.5 +4})\]
\[ b~ln(n)<ln(n^2)-ln({n^.5 +4})\]
\[ b<\frac{ln(n^2)}{ln(n)}-\frac{ln({n^.5 +4})}{ln(n)}\]
\[ b<\lim_{n\to inf}\frac{ln(n^2)}{ln(n)}-\frac{ln({n^.5 +4})}{ln(n)}\]

Answer 3

\[\lim_{n\to inf}\frac{ln(n^2)}{ln(n)}-\frac{ln({n^.5 +4})}{ln(n)}\]
\[\lim_{n\to inf}1+ln(n)-\frac{ln({n^.5 +4})}{ln(n)}\]
after a long and arduous process, b <= 3/2 should suffice