5.5 Alternating series  (Page 4/10)

${\displaystyle \sum }_{n=1}^{\infty }{\left(\mathrm{-1}\right)}^{n+1}\frac{n}{n+3}$

${\displaystyle \sum }_{n=1}^{\infty }{\left(\mathrm{-1}\right)}^{n+1}\frac{\sqrt{n}+1}{\sqrt{n}+3}$

${\displaystyle \sum }_{n=1}^{\infty }{\left(\mathrm{-1}\right)}^{n+1}\frac{1}{\sqrt{n+3}}$

${\displaystyle \sum }_{n=1}^{\infty }{\left(\mathrm{-1}\right)}^{n+1}\frac{\sqrt{n+3}}{n}$

${\displaystyle \sum }_{n=1}^{\infty }{\left(\mathrm{-1}\right)}^{n+1}\frac{1}{n\text{!}}$

${\displaystyle \sum }_{n=1}^{\infty }{\left(\mathrm{-1}\right)}^{n+1}\frac{3^n}{n\text{!}}$

${\displaystyle \sum }_{n=1}^{\infty }{\left(\mathrm{-1}\right)}^{n+1}{\left(\frac{n-1}{n}\right)}^n$

${\displaystyle \sum }_{n=1}^{\infty }{\left(\mathrm{-1}\right)}^{n+1}{\left(\frac{n+1}{n}\right)}^n$

${\displaystyle \sum }_{n=1}^{\infty }{\left(\mathrm{-1}\right)}^{n+1}{\text{sin}}^{2}n$

${\displaystyle \sum }_{n=1}^{\infty }{\left(\mathrm{-1}\right)}^{n+1}{\text{cos}}^{2}n$

${\displaystyle \sum }_{n=1}^{\infty }{\left(\mathrm{-1}\right)}^{n+1}{\text{sin}}^2\left(1\text{/}n\right)$

${\displaystyle \sum }_{n=1}^{\infty }{\left(\mathrm{-1}\right)}^{n+1}{\text{cos}}^2\left(1\text{/}n\right)$

${\displaystyle \sum }_{n=1}^{\infty }{\left(\mathrm{-1}\right)}^{n+1}\text{ln}\left(1\text{/}n\right)$

${\displaystyle \sum }_{n=1}^{\infty }{\left(\mathrm{-1}\right)}^{n+1}\text{ln}\left(1+\frac{1}{n}\right)$

${\displaystyle \sum }_{n=1}^{\infty }{\left(\mathrm{-1}\right)}^{n+1}\frac{n^2}{1+n^4}$

${\displaystyle \sum }_{n=1}^{\infty }{\left(\mathrm{-1}\right)}^{n+1}\frac{n^e}{1+n^{\pi }}$

${\displaystyle \sum }_{n=1}^{\infty }{\left(\mathrm{-1}\right)}^{n+1}{2}^{1\text{/}n}$

${\displaystyle \sum }_{n=1}^{\infty }{\left(\mathrm{-1}\right)}^{n+1}{n}^{1\text{/}n}$

${\displaystyle \sum }_{n=1}^{\infty }{\left(\mathrm{-1}\right)}^n\left(1-n^{1\text{/}n}\right)$ ( Hint: $n^{1\text{/}n}\approx 1+\text{ln}(n)\text{/}n$ for large $n.)$

${\displaystyle \sum }_{n=1}^{\infty }{\left(\mathrm{-1}\right)}^{n+1}n\left(1-\text{cos}\left(\frac{1}{n}\right)\right)$ ( Hint: $\text{cos}(1\text{/}n)\approx 1-1\text{/}{n}^2$ for large $n.)$

${\displaystyle \sum }_{n=1}^{\infty }{\left(\mathrm{-1}\right)}^{n+1}\left(\sqrt{n+1}-\sqrt{n}\right)$ ( Hint: Rationalize the numerator.)

${\displaystyle \sum }_{n=1}^{\infty }{\left(\mathrm{-1}\right)}^{n+1}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)$ ( Hint: Cross-multiply then rationalize numerator.)

${\displaystyle \sum }_{n=1}^{\infty }{\left(\mathrm{-1}\right)}^{n+1}\left(\text{ln}\left(n+1\right)-\text{ln}n\right)$

${\displaystyle \sum }_{n=1}^{\infty }{\left(\mathrm{-1}\right)}^{n+1}n\left({\text{tan}}^{\mathrm{-1}}\left(n+1\right)-{\text{tan}}^{\mathrm{-1}}n\right)$ ( Hint: Use Mean Value Theorem.)

${\displaystyle \sum }_{n=1}^{\infty }{\left(\mathrm{-1}\right)}^{n+1}\left({\left(n+1\right)}^2-n^2\right)$

${\displaystyle \sum }_{n=1}^{\infty }{\left(\mathrm{-1}\right)}^{n+1}\left(\frac{1}{n}-\frac{1}{n+1}\right)$