New Notes Blog: `sum_(n=1)^oo ln((n+1)/n)` Determine the convergence or divergence of the series.

Tuesday, 29 September 2015

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\ln{{\left(\frac{{{n}+{1}}}{{n}}\right)}}}$ Determine the convergence or divergence of the series.

To determine if the series $\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\ln{{\left(\frac{{{n}+{1}}}{{n}}\right)}}}$ converges or diverges, we may apply the Direct Comparison Test.

Direct Comparison test is applicable when $\displaystyle\sum{a}_{{n}}$ and $\displaystyle\sum{b}_{{n}}$ are both positive series for all n where $\displaystyle{a}_{{n}}\leq{b}_{{n}}$ .

If $\displaystyle\sum{b}_{{n}}$ converges then $\displaystyle\sum{a}_{{n}}$ converges.

If $\displaystyle\sum{a}_{{n}}$ diverges so does the $\displaystyle\sum{b}_{{n}}$ diverges.

For the given series $\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\ln{{\left(\frac{{{n}+{1}}}{{n}}\right)}}}$ , we let $\displaystyle{b}_{{n}}={\ln{{\left(\frac{{{n}+{1}}}{{n}}\right)}}}$ .

Let $\displaystyle{a}_{{n}}={\ln{{\left(\frac{{1}}{{n}}\right)}}}$ since $\displaystyle{\ln{{\left(\frac{{1}}{{n}}\right)}}}\leq{\ln{{\left(\frac{{{n}+{1}}}{{n}}\right)}}}$ ...

To determine if the series $\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\ln{{\left(\frac{{{n}+{1}}}{{n}}\right)}}}$ converges or diverges, we may apply the Direct Comparison Test.

Direct Comparison test is applicable when $\displaystyle\sum{a}_{{n}}$ and $\displaystyle\sum{b}_{{n}}$ are both positive series for all n where $\displaystyle{a}_{{n}}\leq{b}_{{n}}$ .

If $\displaystyle\sum{b}_{{n}}$ converges then $\displaystyle\sum{a}_{{n}}$ converges.

If $\displaystyle\sum{a}_{{n}}$ diverges so does the $\displaystyle\sum{b}_{{n}}$ diverges.

For the given series $\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\ln{{\left(\frac{{{n}+{1}}}{{n}}\right)}}}$ , we let $\displaystyle{b}_{{n}}={\ln{{\left(\frac{{{n}+{1}}}{{n}}\right)}}}$ .

Let $\displaystyle{a}_{{n}}={\ln{{\left(\frac{{1}}{{n}}\right)}}}$ since $\displaystyle{\ln{{\left(\frac{{1}}{{n}}\right)}}}\leq{\ln{{\left(\frac{{{n}+{1}}}{{n}}\right)}}}$ .

To evaluate if the series $\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\ln{{\left(\frac{{1}}{{n}}\right)}}}$ converges or diverges, we may apply Divergence test:

$\displaystyle\lim_{{{n}-\gt\infty}}{a}_{{n}}\ne{0}$ or does not exist then the series $\displaystyle\sum{a}_{{n}}$ diverges

We set-up the limit as:

$\displaystyle\lim_{{{n}-\gt\infty}}{\ln{{\left(\frac{{1}}{{n}}\right)}}}=\lim_{{{n}-\gt\infty}}{\ln{{\left({{n}}^{{-{1}}}\right)}}}$

$\displaystyle={\left(-{1}\right)}\lim_{{{n}-\gt\infty}}{\ln{{\left({n}\right)}}}$

$\displaystyle=-\infty$

With the limit value $\displaystyle{L}=-\infty$ , it satisfy $\displaystyle\lim_{{{n}-\gt\infty}}{a}_{{n}}\ne{0}$ .

Thus, the series $\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\ln{{\left(\frac{{1}}{{n}}\right)}}}$ diverges

Conclusion based from Direct Comparison test:

The series $\displaystyle{\sum_{{{n}={1}}}^{\infty}}{a}_{{n}}={\sum_{{{n}={1}}}^{\infty}}{\ln{{\left(\frac{{1}}{{n}}\right)}}}$ diverges then it follows that $\displaystyle{\sum_{{{n}={1}}}^{\infty}}{b}_{{n}}={\sum_{{{n}={1}}}^{\infty}}{\ln{{\left(\frac{{{n}+{1}}}{{n}}\right)}}}$ also diverges.

New Notes Blog

Tuesday, 29 September 2015

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\ln{{\left(\frac{{{n}+{1}}}{{n}}\right)}}}$ Determine the convergence or divergence of the series.

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In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?

Tuesday, 29 September 2015

Determine the convergence or divergence of the series.

No comments:

Post a Comment

In &quot;By the Waters of Babylon,&quot; under the leadership of John, what do you think the Hill People will do with their society?

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\ln{{\left(\frac{{{n}+{1}}}{{n}}\right)}}}$ Determine the convergence or divergence of the series.

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?