Question

Use a comparison test (either normal or limit) to determine whether the following series converges or diverges. Be sure to clearly identify what you are comparing too and if that converges or diverges.

Answer 1

By the comparison test, the series converges.

We have

$\dfrac1{k\sqrt{k+2}} \le \dfrac1{k \sqrt k} = \dfrac1{k^{3/2}}$

so we can compare to a convergent $p$-series,

$\displaystyle \sum_{k=1}^\infty \frac1{k\sqrt{k+2}} \le \sum_{k=1}^\infty \frac1{k^{3/2}} < \infty$

Answer 2

By the comparison test, the series converges.

What are convergence series and diverging series?

If the infinite series converges to a real number it is called converging if not then it is called diverging series.

If the larger series is convergent the smaller series must also be convergent. Likewise, if the smaller series is divergent then the larger series must also be divergent.

For this series to be converging it must follow the following :

⇒ $| \frac{t_{n+1} }{t_{n} } | \leq 1$

⇒Putting n = 1

$| \frac{t_{2} }{t_{1} } | \leq 1$

$\t_{2} t_{1}$ $t_{2}$= 1/4

$t_{1}$=1/$\sqrt{2}$

$\frac{t_{2} }{{t_{1} }}$ = 1/2$\sqrt{2$

⇒ t₂/t₁≤1

⇒ Hence it converges

Learn more about the comparison tests here :

brainly.com/question/12069811

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