Geometric Series
\sum_{n=1}^\infty 2 (\frac{1}{3})^{n-1}
Converges, this is a geometric with
r=\frac{1}{3}.
\text{Converges to } 3.
\sum_{n=1}^\infty \frac{1}{n}
Diverges. This is the harmonic series. OR, you can say this is a p-series with p=1.
\sum_{n=1}^\infty \frac{ln(n)}{n}
The integral
\int_1^\infty \frac{ln(x)}{x} dx
\text{diverges so by the integral test }
\text{ the series diverges.}
\sum_{n=1}^\infty \frac{3}{\sqrt{n}}
By the LCT with a p-series with p=1/2, this series also diverges.
\sum_{n=1}^\infty \frac{n}{n+1}
The terms of this series go to 1, not 0, so the series diverges.
\sum_{n=1}^\infty (\frac{1}{2})^{1-n}
Diverges. This is a geometric series with r=2.
\sum_{n=1}^\infty \frac{1}{n^{2/3}}
Diverges. This is a p-series with p=2/3, which is less than 1.
\sum_{n=1}^\infty e^{-n}
The integral
\int_1^\infty e^{-x} dx
\text{ converges to } e^{-1}
\text{ so the series also converges.}
\sum_{n=1}^\infty \frac{5}{\sqrt{n^8+n^2-7}
By the LCT with a p-series where p=4, this series also converges.
\sum_{n=1}^\infty \frac{2n^2+\cos^2(n)}{5n^9-8}
By the LCT with a p-series where p=7, this series also converges.
\sum_{n=1}^\infty (\frac{5}{4})^{2-n}
Converges, this is geometric with r=4/5. This converges to 25/4.
\sum_{n=1}^\infty \frac{1}{\sqrt{n^7}}
Converges. This is a p-series where p=7/2, which is larger than 1.
\sum_{n=\pi}^\infty sin(n)
This diverges because the integral below does not converge to any number.
\int_\pi^\infty sin(x) dx
\sum_{n=1}^\infty (\frac{3n-1}{4n+3})^n
By the LCT with a geometric series with r=3/4, this series also converges.
\sum_{n=0}^\infty \frac{2^{3-n}}{5^{n+4}}
This is a geometric series with r=1/10 and converges to
\frac{\frac{8}{625}}{1-\frac{1}{10}} = \frac{16}{1125}