Answer to Question #141022 in Calculus for Besmallah Yousefi

Question

Determine whether the following series converges:
S=1-(1/2)-(1/2^2)+(1/2^3)+(1/2^4)-(1/2^5)-(1/2^6)...
1+(1/3sqrt(2))+(1/3sqrt(3)+...+(1/3sqrt(k))...

Answer 1

Given series:

S = 1 - $\frac{1}{2} - \frac{1}{2^2} + \frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}-\frac{1}{2^6}+...$

Let us rearrange the given terms of the series. Then, we get

S = $[1 - \frac{1}{2^2} + \frac{1}{2^4} - \frac{1}{2^6} + ... ] - [\frac{1}{2} - \frac{1}{2^3} + \frac{1}{2^5} + ... ]$

Recollect the following:

$a + ar + ar^2 + ar^3 + ... + ... = \frac{a}{1 - r},$ where $a$ is the first term and $r$ is the common ratio of the infinite geometric series.

Using the above, we have

$S = [\frac{1}{1+\frac{1}{2^2}}] (a = 1, r = \frac{-1}{2^2}) - [\frac{1/2}{1+\frac{1}{2^2}}] (a = 1, r = \frac{-1}{2^2})$

$= \frac{1}{5/4} - \frac{1/2} {5/4}$

$= \frac{4}{5} - \frac{2}{5}$

$= \frac{2}{5}$ (finite value)

Observe that the sum of the given series is finite value.

Therefore, the given series is convergent.

Ans: Convergent

Given series: $1 + \frac{1}{3\sqrt2} + \frac{1}{3\sqrt3} + ... +\frac{1}{3\sqrt k} + ...$

Consider the series from the given series

$\frac{1}{3\sqrt2} + \frac{1}{3\sqrt3} + ... + \frac{1}{3\sqrt k} + ...$

Rewrite the above series as sigma notation.

$\frac{1}{3\sqrt2} + \frac{1}{3\sqrt3} + ... + \frac{1}{3\sqrt k} =\textstyle\sum_2^k\frac{1}{3\sqrt n}$

Recollect the following:

Auxiliary series $\Sigma\frac{1}{n^p}$ is convergent series if $p$ > 1 and divergent if p $\leqslant1$

Using the above, the series $\textstyle\sum_2^k\frac{1}{3\sqrt n}$ is divergent because $p = \frac{1}{2} < 1$

We know that the convergence or divergence of a series is unaltered by adding or deleting some finite number of terms.

Using the above, the series $1 + \textstyle\sum_2^k\frac{1}{3\sqrt n}$ is divergent.

Therefore, the given series is divergent.

Ans: Divergent