Evaluating Summation of an Infinite Series

[jisbon]

Homework Statement

Evaluate $\lim_{n \rightarrow +\infty} \frac {1} {n} [(\frac {1}{n})^{1.5} + (\frac {2}{n})^{1.5} +(\frac {3}{n})^{1.5}+ (\frac {4}{n})^{1.5}+...+(\frac {n}{n})^{1.5}]$

Relevant Equations

NIL

Evaluate $\lim_{n \rightarrow +\infty} \frac {1} {n} [(\frac {1}{n})^{1.5} + (\frac {2}{n})^{1.5} +(\frac {3}{n})^{1.5}+ (\frac {4}{n})^{1.5}+...+(\frac {n}{n})^{1.5}]$

Hello. So I'm solving this question at the moment. I know I'm supposed to find out the summation of this before being able to solve the solution:

$(\frac {1}{n})^{1.5} + (\frac {2}{n})^{1.5} +(\frac {3}{n})^{1.5}+ (\frac {4}{n})^{1.5}+... +(\frac {n}{n})^{1.5}$However, I can't seem to find any AP or GP series in the equation above. There's no fixed difference between $\frac {1}{n}^{1.5}$ and $\frac {2}{n}^{1.5}$ and ... as far as I can tell? Any ideas on how to proceed here? Thank you.

 

[fresh_42]

Can you define the entire series? You cannot expect any answer if only two summands are given. This could be anything.

 

[jisbon]

Can you define the entire series? You cannot expect any answer if only two summands are given. This could be anything.

Edited for clarity. Thank you.

 

[fresh_42]

Edited for clarity. Thank you.

Your parentheses don't match. It looks like $a_n = n^{-\frac{5}{2}}\sum_{k=1}^\infty k^{\frac{3}{2}}$. But every $a_n$ is already infinitely large, so $n$ doesn't matter.

 

[jisbon]

Your parentheses don't match. It looks like $a_n = n^{-\frac{5}{2}}\sum_{k=1}^\infty k^{\frac{3}{2}}$. But every $a_n$ is already infinitely large, so $n$ doesn't matter.

Hi there, I think I might have edited it wrongly. What I meant was something like: $(\frac {1}{n})^{1.5} + (\frac {2}{n})^{1.5} +(\frac {3}{n})^{1.5}+ (\frac {4}{n})^{1.5}+...$

Thanks

 

[fresh_42]

Hi there, I think I might have edited it wrongly. What I meant was something like: $(\frac {1}{n})^{1.5} + (\frac {2}{n})^{1.5} +(\frac {3}{n})^{1.5}+ (\frac {4}{n})^{1.5}+...$

Thanks

This only changes the power of $n$ to $\frac{3}{2}$, but it is still a sum of infinitely large numbers. Each sequence member has a constant value $n$, i.e. a constant factor. The rest is infinitely large. The limit $n \to \infty$ is irrelevant.

 

[jisbon]

This only changes the power of $n$ to $\frac{3}{2}$, but it is still a sum of infinitely large numbers. Each sequence member has a constant value $n$, i.e. a constant factor. The rest is infinitely large. The limit $n \to \infty$ is irrelevant.

Hi, thanks for your reply.
Firstly, my bad for not seeing the question properly, I edited it again as shown
Not sure what you meant by your reply above, but shouldn't I be using a formula for the summation for:
$\sum_{a=1}^{n} (\frac {a}{n})^{1.5}$

 

[fresh_42]

Hi, thanks for your reply.
Firstly, my bad for not seeing the question properly, I edited it again as shown
Not sure what you meant by your reply above, but shouldn't I be using a formula for the summation for:
$\sum_{a=1}^{n} (\frac {a}{n})^{1.5}$

Firstly we have $\left( \frac{k}{n} \right)^{1.5} = \frac{k^{1.5}}{n^{1.5}}$ and secondly a sum $\frac{a}{N}+\frac{b}{N}+\frac{c}{N}+\ldots$ where all terms have the same denominator $N=n^{1.5}$. With the distributive law we have $\frac{1}{n^{1.5}} \cdot (1^{1.5}+2^{1.5}+3^{1.5}+ \ldots)$. The second factor is infinitely large, regardless of the value of $\frac{1}{n^{1.5}}$. But if every sequence member is infinitely large, the limit will be as well.

 

[jisbon]

$\frac {1}{n^{1.5}}⋅(1^{1.5}+2^{1.5}+3^{1.5}+…n^{1.5})$

Ok, so from what I understand, the above equation will sum to 1 eventually? Thanks

 

[fresh_42]

Ok, so from what I understand, the above equation will sum to 1 eventually? Thanks

This is not what I wrote. The sum isn't finite, there are infinitely many terms which sum up to infinity. The factor $\frac{1}{n^{1.5}}$ doesn't change that.

 

[jisbon]

This is not what I wrote. The sum isn't finite, there are infinitely many terms which sum up to infinity. The factor $\frac{1}{n^{1.5}}$ doesn't change that.

Okay I think I do understand now. Since $\frac {1}{n^{1.5}}⋅(1^{1.5}+2^{1.5}+3^{1.5}+…n^{1.5})$ =$\infty$,
$\lim_{n \rightarrow +\infty} \frac {1} {n} [(\frac {1}{n})^{1.5} + (\frac {2}{n})^{1.5} +(\frac {3}{n})^{1.5}+ (\frac {4}{n})^{1.5}+...+(\frac {n}{n})^{1.5}]$ =
$\lim_{n \rightarrow +\infty} \frac {1} {n} [\infty]?$

 

[fresh_42]

Sorry, you have edited your question so often that I missed when you turned an infinite sum into a finite one. I thought you meant $\sum_{k=1}^\infty k^{1.5}$.

If the sum is finite, things are different. Currently you have
$$
a_n = \frac{1}{n} \cdot \frac{1}{n^{1.5}} \cdot \left( 1^{1.5}+ 2^{1.5}+\ldots +n^{1.5} \right)= \frac{1}{n^{2.5}} \sum_{k=1}^n k^{1.5}
$$
I don't know a formula for $\sum_{k=1}^n k^{1.5}$, but it is obviously less than $n \cdot n^{1.5}=n^{2.5}$, so $1$ is an upper bound for your limit. There are probably approximations for the sum. If it behaves like $n^{2.5-c}$ for some $c$ independent of $n$, then the limit $n \to \infty$ will be zero. So all depends on how $\sum_{k=1}^n k^{1.5}$ behaves.

 

[Dick]

It's not hard to get an approximate formula for $\sum_{k=1}^n k^{1.5}$. Just replace the sum with an integral. You should also be able to get a subleading term by approximating the error you make by doing that.

 

[jisbon]

It's not hard to get an approximate formula for $\sum_{k=1}^n k^{1.5}$. Just replace the sum with an integral. You should also be able to get a subleading term by approximating the error you make by doing that.

Sorry, you have edited your question so often that I missed when you turned an infinite sum into a finite one. I thought you meant $\sum_{k=1}^\infty k^{1.5}$.

If the sum is finite, things are different. Currently you have
$$
a_n = \frac{1}{n} \cdot \frac{1}{n^{1.5}} \cdot \left( 1^{1.5}+ 2^{1.5}+\ldots +n^{1.5} \right)= \frac{1}{n^{2.5}} \sum_{k=1}^n k^{1.5}
$$
I don't know a formula for $\sum_{k=1}^n k^{1.5}$, but it is obviously less than $n \cdot n^{1.5}=n^{2.5}$, so $1$ is an upper bound for your limit. There are probably approximations for the sum. If it behaves like $n^{2.5-c}$ for some $c$ independent of $n$, then the limit $n \to \infty$ will be zero. So all depends on how $\sum_{k=1}^n k^{1.5}$ behaves.

Am I supposed to use the Integer Power Sum Formula here?
If so,
Is $\sum_{k=1}^n k^{1.5}$ simply = $\frac {n^{2.5} + (n+1)^{2.5}}{5}$?

 

[WWGD]

Notice this can be seen as a Riemann sum. Just need to find which and then compute the integral.

 

[Dick]

Am I supposed to use the Integer Power Sum Formula here?
If so,
Is $\sum_{k=1}^n k^{1.5}$ simply = $\frac {n^{2.5} + (n+1)^{2.5}}{5}$?

Yes, that's what you get approximating by integrals. And it's not an equality, it's only a pretty good approximation to the sum.

 

[jisbon]

Yes, that's what you get approximating by integrals. And it's not an equality, it's only a pretty good approximation to the sum.

So by evaluating it can I take the answer as the approximation? Or is there an accurate way to determine the value?

Notice this can be seen as a Riemann sum. Just need to find which and then compute the integral.

I searched up on Riemann sum and apparently it is a certain kind of approximation of an integral by a finite sum. This means I will integrate the graph $y= x^1.5$ from x=1 to infinity?

 

[Dick]

So by evaluating it can I take the answer as the approximation? Or is there an accurate way to determine the value?

I searched up on Riemann sum and apparently it is a certain kind of approximation of an integral by a finite sum. This means I will integrate the graph $y= x^1.5$ from x=1 to infinity?

The Riemann sum is how you derive your power sum formula. If you draw your power sum as rectangles they are an upper sum for the curve $y=x^{1.5}$ and a lower sum for the curve $y=(x+1)^{1.5}$. You integrate up to $n$ and average the two to get your formula. It's plenty accurate enough to use to figure the limit. Follow fresh_42's clues. If you want to be rigorous you can note that the true sum is between the area of the two curves.