1. ## Clarifying another problem

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.

2. 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.

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

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

4. 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

5. 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?

6. 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