Question

Determine if diverges, converges, or converges conditionally.

Answer 1

The given series is conditionally convergent. This can be obtained by using alternating series test first and then comparing the series to the harmonic series.

Determine if diverges, converges, or converges conditionally:

Initially we need to know what Absolute convergence and Conditional convergence,

If $\sum|a_{n} |$ → converges, and $\sum a_{n}$ → converges, then the series is Absolute convergence

If $\sum|a_{n} |$ → diverges, and $\sum a_{n}$ → converges, then the series is Conditional convergence

First use alternating series test,

$\lim_{k \to \infty} \frac{k^{5} +1}{k^{6}+11 }$ = $\lim_{n \to \infty} \frac{5}{6k}$ = 0,

The series is a positive, decreasing sequence that converges to 0.

Next by comparing the series to harmonic series,

$\sum^{\infty} _{k=2}|(-1)^{k+1} \frac{k^{5} +1}{k^{6}+11 }|=\sum^{\infty} _{k=2}\frac{k^{5} +1}{k^{6}+11 }$ ≈  $\sum^{\infty} _{k=2}\frac{1}{k}$ = 0

This implies that the series is divergent by comparison to the harmonic series.

First we got that the series is converging and then we got the series is divergent. Therefore the series is conditionally convergent.

$\sum|a_{n} |$ → diverges, and $\sum a_{n}$ → converges, then the series is Conditional convergence.

Hence the given series is conditionally convergent.

Learn more about conditionally convergent here:

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Answer 2

By the alternating series test, this series converges: for $k\in\Bbb N$,

$\dfrac{k^5+1}{k^6+11}$

is a positive, decreasing sequence that converges to 0.

However,

$\displaystyle \sum_{k=2}^\infty \left| (-1)^{k+1} \frac{k^5+1}{k^6+11} \right| = \sum_{k=2}^\infty \frac{k^5+1}{k^6+11} \approx \sum_{k=2}^\infty \frac1k$

is a divergent series by comparison to the harmonic series.

So the given series is conditionally convergent.