Question

Evaluate the following series:

Answer 1

By the limit comparison test, the expression √[1 / (1 + 1 / k)] - √[1 / (1 + 3 / k)] has a limit, then the expression [1 / √(k + 1)] / [1 /√k] -  [1 / √(k + 3)] / [1 /√k] has a limit and the series ∑ [1 / √(k + 1)] - ∑ [1 / √(k + 3)] is convergent.

Is the series convergent?

Herein we have a series that involves radical components. First, we simplify the expression given:

∑ [1 / √(k + 1) - 1 / √(k + 3)] = ∑ [1 / √(k + 1)] - ∑ [1 / √(k + 3)]

The convergence of the series can be proved by the limit comparison test, where each component of the subtraction of the series is compared with a series that is convergent. We notice that both 1 / √(k + 1) and 1 / √(k + 3) resembles the expresion 1 /√k. Then, we have the following subtraction of ratios:

[1 / √(k + 1)] / [1 /√k] - [1 / √(k + 3)] / [1 /√k]

√k / √(k + 1) - √k / √(k + 3)

√[k / (k + 1)] - √[k / (k + 3)]

Then, by using the limit property for rational functions we find the following result for n → + ∞:

√[1 / (1 + 0)] - √[1 / (1 + 0)]

√1 - √1

1 - 1

0

By the limit comparison test, the expression √[1 / (1 + 1 / k)] - √[1 / (1 + 3 / k)] has a limit, then the expression [1 / √(k + 1)] / [1 /√k] -  [1 / √(k + 3)] / [1 /√k] has a limit and the series ∑ [1 / √(k + 1)] - ∑ [1 / √(k + 3)] is convergent.

Remark

The statement is incomplete and complete form cannot be found, therefore, we decided to determine if the series is convergent or not.

To learn more on convergence: brainly.com/question/15415793

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Answer 2

This is a telescoping sum. The K-th partial sum is

$S_K = \displaystyle \sum_{k=1}^K \left(\frac1{\sqrt{k+1}} - \frac1{\sqrt{k+3}}\right) \\\\ ~~~= \left(\frac1{\sqrt2} - \frac1{\sqrt4}\right) + \left(\frac1{\sqrt3} - \frac1{\sqrt5}\right) + \left(\frac1{\sqrt4} - \frac1{\sqrt6}\right) + \left(\frac1{\sqrt5} - \frac1{\sqrt7}\right) + \cdots \\\\ ~~~~~~~~+ \left(\frac1{\sqrt{K-1}} - \frac1{\sqrt{K+1}}\right) \\\\ ~~~~~~~~+ \left(\frac1{\sqrt K} - \frac1{\sqrt{K+2}}\right) + \left(\frac1{\sqrt{K+1}} - \frac1{\sqrt{K+3}}\right)$

$\displaystyle = \frac1{\sqrt2} + \frac1{\sqrt3} - \frac1{\sqrt{K+2}} - \frac1{\sqrt{K+3}}$

As $K\to\infty$, the two trailing terms will converge to 0, and the overall infinite sum will converge to

$\displaystyle \sum_{k=1}^\infty \left(\frac1{\sqrt{k+1}} - \frac1{\sqrt{k+3}}\right) = \lim_{k\to\infty} S_k = \boxed{\frac1{\sqrt2} + \frac1{\sqrt3}}$