Clarifying another problem

**auslmar** · July 5th 2008, 06:55 PM

Hello,

I'm having trouble interpreting anothher problem. There is a series $a_n = \sum_{n=1}^{\infty}\frac{(5n^\frac{7}{6}+9)}{(8n^9 +4n^6+4*n^3+2)^\frac{1}{2}}$

$a_n$ converges, is a series to which the limit comparison test applies with comparing series a p-series, and is a series to which the limit comparison test applies with comparing series with terms $n^pr^n$ , i.e. $\sum_{n=1}^{\infty}n^pr^n$ , for some $p$ and $r$ with $r > 0$ .

"If the series converges, give the limit within one percent. If the limit does not exist answer infinity, -infinity or DNE as appropriate"

Well, it is given that the series converges, but I don't understand what is being asked by finding the limit within one percent. Find the sum of the series or what? I don't understand how to find the sum of the series, either. If anyone has any ideas as to how I should approach this, I would greatly appreciate it.

Thanks for your help.

**CaptainBlack** · July 5th 2008, 11:29 PM

Originally Posted by auslmar

Hello,

I'm having trouble interpreting anothher problem. There is a series $a_n = \sum_{n=1}^{\infty}\frac{(5n^\frac{7}{6}+9)}{(8n^9 +4n^6+4*n^3+2)^\frac{1}{2}}$

$a_n$ converges, is a series to which the limit comparison test applies with comparing series a p-series, and is a series to which the limit comparison test applies with comparing series with terms $n^pr^n$ , i.e. $\sum_{n=1}^{\infty}n^pr^n$ , for some $p$ and $r$ with $r > 0$ .

"If the series converges, give the limit within one percent. If the limit does not exist answer infinity, -infinity or DNE as appropriate"

Well, it is given that the series converges, but I don't understand what is being asked by finding the limit within one percent. Find the sum of the series or what? I don't understand how to find the sum of the series, either. If anyone has any ideas as to how I should approach this, I would greatly appreciate it.

Thanks for your help.

You are expected to sum the first few terms, going as far as you need to ensure the error in the partial sum is less than 1% of the true sum.

(This should be about 4 terms)

RonL

**auslmar** · July 6th 2008, 12:54 AM

But, how would I sum them? Just add $a_1 + a_2 + a_3 + a_4$ ?

**CaptainBlack** · July 6th 2008, 01:19 AM

Originally Posted by auslmar

But, how would I sum them? Just add $a_1 + a_2 + a_3 + a_4$ ?

Sorry there is some confusion here, you have:

$ a_n = \sum_{n=1}^{\infty}\frac{(5n^\frac{7}{6}+9)}{(8n^9 +4n^6+4n^3+2)^\frac{1}{2}} $ ,

but $a_n$ is not dependent on $n$ , so I have assumed you mean:

$ a_N = \sum_{n=1}^{N}\frac{(5n^\frac{7}{6}+9)}{(8n^9+4n^6 +4n^3+2)^\frac{1}{2}} $

and I am talking about the convergence of $a_N$ to $a$ as $N \to \infty$ , and to getting this to a fractional error of less than $0.01$ (or 1%). For this you will need about $N=4$ .

RonL

**auslmar** · July 6th 2008, 11:13 AM

Originally Posted by CaptainBlack

Sorry there is some confusion here, you have:

$ a_n = \sum_{n=1}^{\infty}\frac{(5n^\frac{7}{6}+9)}{(8n^9 +4n^6+4n^3+2)^\frac{1}{2}} $ ,

but $a_n$ is not dependent on $n$ , so I have assumed you mean:

$ a_N = \sum_{n=1}^{N}\frac{(5n^\frac{7}{6}+9)}{(8n^9+4n^6 +4n^3+2)^\frac{1}{2}} $

and I am talking about the convergence of $a_N$ to $a$ as $N \to \infty$ , and to getting this to a fractional error of less than $0.01$ (or 1%). For this you will need about $N=4$ .

RonL

Well, I can see what is meant now, and it turns out that this led me to the correct answer. I do have a follow-up question, though: how did you determine that it would take 4?

**CaptainBlack** · July 6th 2008, 01:00 PM

Originally Posted by auslmar

Well, I can see what is meant now, and it turns out that this led me to the correct answer. I do have a follow-up question, though: how did you determine that it would take 4?

Well, the asymtotic form for the n-th term is $\sim 5n^{-10/3}$ , and if we sum a few terms we find the sum $\approx 4$ , so we need the error to be less than $\approx 0.03$ (that is 1% of 4 rounded down as the sum is slightly less than 4, and we need this allowable error to be underestimated to compensate for some of the approximations).

We have a rule of thumb that the error in truncating the series is of the same order as the first neglected term, so to find the number of terms needed we solve:

$0.03=5 (n+1)^{-10/3}$

and round up to the next larger integer.

This is an approximate method rather than exact, but in this case is good enough.

RonL

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