Blog: The ratio test to solve

The series in the problem is $\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{{2}+{{\left(-{1}\right)}}^{{n}}}}{{{1.25}}^{{n}}}$

To determine the convergence of the series let us use the comparison test. If there are two series $\displaystyle{S}_{{a}}$ and $\displaystyle{S}_{{b}}$ with terms $\displaystyle{T}_{{a}}$ and $\displaystyle{T}_{{b}}$ and $\displaystyle{T}_{{b}}\ge{T}_{{a}}$ , if the series $\displaystyle{S}_{{b}}$ converges, $\displaystyle{S}_{{a}}$ also converges.

Now, $\displaystyle{{\left(-{1}\right)}}^{{n}}$ can be either positive or negative and this is dependent on whether n is even or odd.

Let $\displaystyle{T}_{{b}}=\frac{{{2}+{{1}}^{{n}}}}{{{1.25}}^{{n}}}$ . If...

The series in the problem is $\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{{2}+{{\left(-{1}\right)}}^{{n}}}}{{{1.25}}^{{n}}}$

Now, $\displaystyle{{\left(-{1}\right)}}^{{n}}$ can be either positive or negative and this is dependent on whether n is even or odd.

Let $\displaystyle{T}_{{b}}=\frac{{{2}+{{1}}^{{n}}}}{{{1.25}}^{{n}}}$ . If $\displaystyle{T}_{{a}}=\frac{{{2}+{{\left(-{1}\right)}}^{{n}}}}{{{1.25}}^{{n}}}$ , $\displaystyle{T}_{{b}}\ge{T}_{{a}}$

To determine the convergence of $\displaystyle{S}_{{b}}$ we use the ratio test.

First we find the value of L = $\displaystyle\lim_{{{n}=\infty}}{\left|\frac{{T}_{{{n}+{1}}}}{{T}_{{n}}}\right|}$

$\displaystyle{T}_{{{n}+{1}}}=\frac{{{2}+{{\left({1}\right)}}^{{{n}+{1}}}}}{{{1.25}}^{{{n}+{1}}}}$

= $\displaystyle\frac{{{2}+{{1}}^{{n}}}}{{{1.25}\cdot{{1.25}}^{{n}}}}$

L = $\displaystyle\lim_{{{n}=\infty}}{\left|\frac{{T}_{{{n}+{1}}}}{{T}_{{n}}}\right|}$

= $\displaystyle\lim_{{{n}=\infty}}{\left|\frac{{\frac{{{2}+{{1}}^{{n}}}}{{{1.25}\cdot{{1.25}}^{{n}}}}}}{{\frac{{{2}+{{1}}^{{n}}}}{{{1.25}}^{{n}}}}}\right|}$

= $\displaystyle\frac{{1}}{{1.25}}$

As 1.25 is greater than 1, $\displaystyle\frac{{1}}{{1.25}}$ is less than 1.

By the ratio test, the series $\displaystyle{S}_{{b}}=\sum\frac{{{2}+{{1}}^{{n}}}}{{{1.25}}^{{n}}}$ converges.

As $\displaystyle{S}_{{b}}$ converges, the series $\displaystyle{S}_{{a}}=\sum\frac{{{2}+{{\left(-{1}\right)}}^{{n}}}}{{{1.25}}^{{n}}}$ also converges.

The given series $\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{{2}+{{\left(-{1}\right)}}^{{n}}}}{{{1.25}}^{{n}}}$ converges.

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The ratio test to solve

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The ratio test to solve

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