# Integral Test/Comparison Test

• Jan 27th 2009, 11:51 AM
Integral Test/Comparison Test
infinity over Sigma n=1 n^2/n^3 + 1 Use the integral test to see if the infinite series converges.

infinity over Sigma n=2 n^2+1/n^3.5-2 Use either the Comparison Test or the Limit Comparison Test to see if the infinite series converges.
• Jan 27th 2009, 11:58 AM
Krizalid
Quote:

Put $f(x)=\frac{{{x}^{3}}}{{{x}^{2}}+1},$ then $f$ is a positive, continuous and decreasing function for $x\ge2,$ hence, the integral test applies and the series will converge or diverge if $\int_{2}^{\infty }{\frac{{{x}^{3}}}{{{x}^{2}}+1}\,dx}$ does.
Since $\frac{{{x}^{3}}}{{{x}^{2}}+1}\ge \frac{{{x}^{3}}}{{{x}^{2}}+{{x}^{2}}}=\frac{1}{2}x ,$ then the integral diverges, so does the series. (Actually, one can bound immediately the general term of the series rather than applying the integral test.)
As for your second question, compare the series with $\sum\limits_{n=2}^{\infty }{\frac{1}{{{n}^{1,5}}}}$ which is a convergent $p-$series with $p=1,5>1.$
So, put ${{a}_{n}}=\frac{{{n}^{2}}+1}{{{n}^{3,5}}-2}$ and ${{b}_{n}}=\frac{1}{{{n}^{1,5}}},$ and observe that $\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{a}_{n}}}{{{b}_{n}}}=1,$ so your series converges.