Solving Series Convergence Problems: 1+ \frac{1}{3}-\frac{1}{2}+\frac{1}{5}+...

[Lisa91]

I have a problem with convergence of two series:

1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}+...

1+ \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{5}}+\frac{1}{ \sqrt{6}}-\frac{1}{\sqrt{7}}-\frac{1}{\sqrt{8}}+...

Could you give me please any hints so that I can solve them?

 

[chisigma]

I have a problem with convergence of two series:

1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}+...

1+ \frac{1}{sqrt{2}}-\frac{1}{sqrt{3}}-\frac{1}{sqrt{4}}+\frac{1}{sqrt{5}}+\frac{1}{sqrt{6}}-\frac{1}{sqrt{7}}-\frac{1}{sqrt{8}}+...

Could you give me please any hints so that I can solve them?

The first series is a classical example of the properties of a conditionally convergent series. We start with the well known series... $\displaystyle 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} -... = \ln 2$ (1)

... from which we derive...

$\displaystyle \frac{1}{2} - \frac{1}{4} + \frac{1}{6} - \frac{1}{8} + \frac{1}{10} - \frac{1}{12} + \frac{1}{14} -... = \frac{\ln 2}{2}$ (2)

... that can be written as...

$\displaystyle 0 + \frac{1}{2} + 0 - \frac{1}{4} + 0 + \frac{1}{6}+ 0 - \frac{1}{8} + 0 + \frac{1}{10}+ 0 - \frac{1}{12} + 0 + \frac{1}{14} +... = \frac{\ln 2}{2}$ (3)

The we cas sum 'term by term' (1) and (3) obtaining...

$\displaystyle 1 - \frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} - \frac{1}{4} + \frac{1}{9} + \frac{1}{11} - \frac{1}{6} + ... = \frac{3}{2}\ \ln 2$ (4)

Kind regards

$\chi$ $\sigma$

 

[Lisa91]

May I write
1- \frac{1}{sqrt{2}}+\frac{1}{sqrt{3}}-\frac{1}{sqrt{4}}+\frac{1}{sqrt{5}}-\frac{1}{sqrt{6}}+\frac{1}{sqrt{7}}-... = (\lnn)^{\frac{1}{2}}

 

[chisigma]

I have a problem with convergence of two series:

1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}+...

1+ \frac{1}{sqrt{2}}-\frac{1}{sqrt{3}}-\frac{1}{sqrt{4}}+\frac{1}{sqrt{5}}+\frac{1}{sqrt{6}}-\frac{1}{sqrt{7}}-\frac{1}{sqrt{8}}+...

Could you give me please any hints so that I can solve them?

The second series can be written in the form...

$\displaystyle \sum_{n=0}^{\infty} (-1)^{n} a_{n}\ ,\ a_{n}= \frac{1}{\sqrt{2 n + 1}} + \frac{1}{\sqrt{2 n + 2}}$ (1)

Now is $a_{n+1}<a_{n}$ and $\lim_{n \rightarrow \infty} a_{n}=0$ so that for the Leibnitz's criterion the series converges...

Kind regards

$\chi$ $\sigma$

 

[MarkFL]

Just a few $\displaystyle LaTeX$ suggestions:

To express the square root of a value, use the code \sqrt{x}, and for the nth root, use \sqrt[n]{x}.

Your natural log function on the right side is rendered incorrectly because there is no space between it and its argument. I suggest the code \ln(n).

 

[chisigma]

May I write
1- \frac{1}{sqrt{2}}+\frac{1}{sqrt{3}}-\frac{1}{sqrt{4}}+\frac{1}{sqrt{5}}-\frac{1}{sqrt{6}}+\frac{1}{sqrt{7}}-... = (\lnn)^{\frac{1}{2}}

Is...

$\displaystyle 1- \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{6}}+\frac{1}{\sqrt{7}}-...= (1-\sqrt{2})\ \zeta(\frac{1}{2}) = .6048986434...$ (1)

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Because is...

$\displaystyle 1- \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{6}}+\frac{1}{\sqrt{7}}-...= (1-\sqrt{2})\ \zeta(\frac{1}{2})$ (1)

... where $\zeta(*)$ is the Riemann Zeta Function, it is also...

$\displaystyle \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{4}}+\frac{1}{\sqrt{6}}-\frac{1}{\sqrt{8}}+\frac{1}{\sqrt{10}}-\frac{1}{\sqrt{12}}+\frac{1}{\sqrt{14}}-...= \frac{1-\sqrt{2}}{\sqrt{2}}\ \zeta(\frac{1}{2})$ (2)

Now remembering the definition of Diriclet Beta Function...

$\displaystyle \beta(s)= \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n + 1)^{s}}$ (3)

... we obtain...

$\displaystyle 1 + \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{4}} + \frac{1}{\sqrt{5}} + \frac{1}{\sqrt{6}} - ... = \frac{1-\sqrt{2}}{\sqrt{2}}\ \zeta(\frac{1}{2}) + \beta(\frac{1}{2})$ (4)

Kind regards

$\chi$ $\sigma$