New Notes Blog: `sum_(n=1)^oo 1/(nsqrt(n^2+1))` Use the Limit Comparison Test to determine the convergence or divergence of the series.

Saturday, 4 July 2015

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{1}}{{{n}\sqrt{{{{n}}^{{2}}+{1}}}}}$ Use the Limit Comparison Test to determine the convergence or divergence of the series.

Recall Limit Comparison Test considers two positive series $\displaystyle{a}_{{n}}\geq{0}$ and $\displaystyle{b}_{{n}}\gt{0}$ for all $\displaystyle{n}$ such that the limit from the ratio of two series as:

$\displaystyle\lim_{{{n}-\gt\infty}}\frac{{a}_{{n}}}{{b}_{{n}}}={c}$

where $\displaystyle{c}$ is positive and finite $\displaystyle{\left({0}\lt{c}\lt\infty\right)}$ .

When we satisfy the condition for the limit value, the two series will have the same properties. Both will either converge or diverges. We may also consider the conditions:

If we have $\displaystyle\lim_{{{n}-\gt\infty}}\frac{{a}_{{n}}}{{b}_{{n}}}={0}$ , we follow: $\displaystyle\sum{b}_{{n}}$ converges then $\displaystyle\sum{a}_{{n}}$ converges.

If we have $\displaystyle\lim_{{{n}-\gt\infty}}\frac{{a}_{{n}}}{{b}_{{n}}}=\infty$ , we follow: $\displaystyle\sum{b}_{{n}}$ diverges then $\displaystyle\sum{a}_{{n}}$ diverges.

For the given series $\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{1}}{{{n}\sqrt{{{{n}}^{{2}}+{1}}}}}$ , we may let $\displaystyle{a}_{{n}}=\frac{{1}}{{{n}\sqrt{{{{n}}^{{2}}+{1}}}}}.$

Rationalize the denominator:

$\displaystyle\frac{{1}}{{{n}\sqrt{{{{n}}^{{2}}+{1}}}}}\cdot\frac{\sqrt{{{{n}}^{{2}}+{1}}}}{\sqrt{{{{n}}^{{2}}+{1}}}}=\frac{\sqrt{{{{n}}^{{2}}+{1}}}}{{{n}{\left({{n}}^{{2}}+{1}\right)}}}$

Note: $\displaystyle\sqrt{{{{n}}^{{2}}+{1}}}\cdot\sqrt{{{{n}}^{{2}}+{1}}}={{\left(\sqrt{{{{n}}^{{2}}+{1}}}\right)}}^{{2}}={{n}}^{{2}}+{1}$ .

Ignoring the constants, we get:

$\displaystyle\frac{\sqrt{{{{n}}^{{2}}+{1}}}}{{{n}{\left({{n}}^{{2}}+{1}\right)}}}\approx\frac{\sqrt{{{{n}}^{{2}}}}}{{{n}{\left({{n}}^{{2}}\right)}}}{\quad\text{or}\quad}\frac{{1}}{{{n}}^{{2}}}$

Note: $\displaystyle\sqrt{{{{n}}^{{2}}}}={n}$ . We may cancel it out to simplify.

This gives us a hint that we may apply comparison between the two series: $\displaystyle{a}_{{n}}=\frac{{1}}{{{n}\sqrt{{{{n}}^{{2}}+{1}}}}}$ and $\displaystyle{b}_{{n}}=\frac{{1}}{{{n}}^{{2}}}$ .

The limit of the ratio of the two series will be:

$\displaystyle\lim_{{{n}-\gt\infty}}\frac{{\frac{{1}}{{{n}\sqrt{{{{n}}^{{2}}+{1}}}}}}}{{\frac{{1}}{{{n}}^{{2}}}}}=\lim_{{{n}-\gt\infty}}\frac{{1}}{{{n}\sqrt{{{{n}}^{{2}}+{1}}}}}\cdot\frac{{{n}}^{{2}}}{{1}}$

$\displaystyle=\lim_{{{n}-\gt\infty}}\frac{{{n}}^{{2}}}{{{n}\sqrt{{{{n}}^{{2}}+{1}}}}}$

$\displaystyle=\lim_{{{n}-\gt\infty}}\frac{{n}}{\sqrt{{{{n}}^{{2}}+{1}}}}$

Apply algebraic techniques to evaluate the limit. We divide by n with the highest exponent which is $\displaystyle{n}$ or $\displaystyle{{n}}^{{1}}$ . Note: $\displaystyle{n}$ is the same as $\displaystyle\sqrt{{{{n}}^{{2}}}}$ .

$\displaystyle\lim_{{{n}-\gt\infty}}\frac{{n}}{{\sqrt{{{{n}}^{{2}}+{1}}}}}=\lim_{{{n}-\gt\infty}}\frac{{\frac{{n}}{{n}}}}{{\frac{\sqrt{{{{n}}^{{2}}+{1}}}}{\sqrt{{{{n}}^{{2}}}}}}}$

$\displaystyle=\lim_{{{n}-\gt\infty}}\frac{{1}}{\sqrt{{{1}+\frac{{1}}{{{n}}^{{2}}}}}}$

$\displaystyle=\frac{{1}}{\sqrt{{{1}+\frac{{1}}{\infty}}}}$

$\displaystyle=\frac{{1}}{\sqrt{{{1}+{0}}}}$

$\displaystyle=\frac{{1}}{\sqrt{{{1}}}}$

$\displaystyle=\frac{{1}}{{1}}$

$\displaystyle={1}$

The limit value $\displaystyle{c}={1}$ satisfies $\displaystyle{0}\lt{c}\lt\infty$ .

Apply the p-series test: $\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{1}}{{{n}}^{{p}}}$ is convergent if $\displaystyle{p}\gt{1}$ and divergent if $\displaystyle{p}\leq{1}$ .

The $\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{1}}{{{n}}^{{2}}}$ has $\displaystyle{p}={2}$ which satisfy $\displaystyle{p}\gt{1}$ since $\displaystyle{2}\gt{1}$ . Then, the series $\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{1}}{{{n}}^{{2}}}$ is convergent.

Conclusion based from limit comparison test:

With the series $\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{1}}{{{n}}^{{2}}}$ convergent, it follows the series $\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{1}}{{{n}\sqrt{{{{n}}^{{2}}+{1}}}}}$ is also convergent.

New Notes Blog

Saturday, 4 July 2015

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{1}}{{{n}\sqrt{{{{n}}^{{2}}+{1}}}}}$ Use the Limit Comparison Test to determine the convergence or divergence of the series.

No comments:

Post a Comment

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

Saturday, 4 July 2015

Use the Limit Comparison Test to 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?

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{1}}{{{n}\sqrt{{{{n}}^{{2}}+{1}}}}}$ Use the Limit Comparison Test to 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?