Convergence of Series with Square Roots

[shamieh]

Determine whether the following series converge or diverge. You may use any appropriate test provided you explain your answer.

\(\displaystyle \sum^{\infty}_{n = 1} (\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n + 1}})\)

So it looks like this problem wants me to do telescoping series or something of that nature but I really want to avoid writing out those terms, can't I just split this into two sums and say this:

\(\displaystyle \sum^{\infty}_{n = 1} \frac{1}{\sqrt{n}} - \sum^{\infty}_{n = 1}\frac{1}{\sqrt{n + 1}}\)

Evaluating both of the series using p series test since, p \(\displaystyle \le\) 1 for both series then the whole series diverges?

Is this correct?

 

[Ackbach]

Determine whether the following series converge or diverge. You may use any appropriate test provided you explain your answer.

\(\displaystyle \sum^{\infty}_{n = 1} (\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n + 1}})\)

So it looks like this problem wants me to do telescoping series or something of that nature but I really want to avoid writing out those terms,

If I were you, I would try to curb your distaste for writing out those terms. ;)

can't I just split this into two sums and say this:

\(\displaystyle \sum^{\infty}_{n = 1} \frac{1}{\sqrt{n}} - \sum^{\infty}_{n = 1}\frac{1}{\sqrt{n + 1}}\)

Evaluating both of the series using p series test since, p \(\displaystyle \le\) 1 for both series then the whole series diverges?

Is this correct?

No, because you run into $\infty-\infty$ issues (that subtraction is undefined).

 

[shamieh]

So I can't use p series at all? Are you sure about that?

 

[Ackbach]

So I can't use p series at all? Are you sure about that?

Pretty sure! And I do think the original series converges, so the p series test simply doesn't apply in this situation.

 

[shamieh]

So would you suggest I write this out as a telescoping series and see what I'm left with?

 

[Prove It]

Determine whether the following series converge or diverge. You may use any appropriate test provided you explain your answer.

\(\displaystyle \sum^{\infty}_{n = 1} (\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n + 1}})\)

So it looks like this problem wants me to do telescoping series or something of that nature but I really want to avoid writing out those terms, can't I just split this into two sums and say this:

\(\displaystyle \sum^{\infty}_{n = 1} \frac{1}{\sqrt{n}} - \sum^{\infty}_{n = 1}\frac{1}{\sqrt{n + 1}}\)

Evaluating both of the series using p series test since, p \(\displaystyle \le\) 1 for both series then the whole series diverges?

Is this correct?

I'd write down what the finite sum (up to N terms) would be and then seeing what happens as $\displaystyle \begin{align*} N \to \infty \end{align*}$

$\displaystyle \begin{align*} \sum_{n = 1}^N{ \left( \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}} \right) } &= \left( \frac{1}{\sqrt{1}} - \frac{1}{\sqrt{2}} \right) + \left( \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}} \right) + \left( \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{4}} \right) + \dots + \left( \frac{1}{\sqrt{N}} - \frac{1}{\sqrt{N + 1} } \right) \\ &= \frac{1}{\sqrt{1}} - \frac{1}{\sqrt{N + 1}} \\ &= 1 - \frac{1}{\sqrt{N + 1}} \end{align*}$

and so

$\displaystyle \begin{align*} \sum_{n = 1}^{\infty}{ \left( \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n + 1}} \right) } &= \lim_{N \to \infty}\sum_{n = 1}^N{ \left( \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n + 1}} \right) } \\ &= \lim_{N \to \infty} \left( 1 - \frac{1}{\sqrt{N + 1}} \right) \\ &= 1 - 0 \\ &= 1 \end{align*}$

The sum is clearly convergent (and actually can be evaulated).

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Actually, no, the series diverges. So how are you getting converges? Looks like the p series will indeed work then?

Look.
https://www.wolframalpha.com/input/?i=summation++++1/sqrt(n)+-+1/(sqrt(n)+++1)

It would help if you write out the series correctly. You have put $\displaystyle \begin{align*} \sum_{ n = 1}^{\infty}{ \left( \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n} + 1} \right) } \end{align*}$ when it should be $\displaystyle \begin{align*} \sum_{n = 1}^{\infty}{ \left( \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n + 1}} \right) } \end{align*}$. The series is convergent.

 

[shamieh]

It would help if you write out the series correctly. You have put $\displaystyle \begin{align*} \sum_{ n = 1}^{\infty}{ \left( \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n} + 1} \right) } \end{align*}$ when it should be $\displaystyle \begin{align*} \sum_{n = 1}^{\infty}{ \left( \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n + 1}} \right) } \end{align*}$. The series is convergent.

Yes that would help now wouldn't it?

 

[shamieh]

Thank you for typing that out, I found that same result. Do I need to say that the series converges to 1 or just simply that it is convergent?

 

[Prove It]

It depends on what the question is asking. If it asks if the series is convergent, just say it's convergent. If it asks what the sum converges to, say it.