Finding a partial sum, then estimating the error from the total sum of a series.

[marek]

"Find the 5th partial sum, and using an error formula, estimate the error of this sum with respect to the sum of the series."

The summation from n=1 to infinity of [(-1)^n]/n!

I evaluated the partial sum through a calculator, my answer was -19/30, or -.63333333


I'm having difficulty estimating how far this partial sum is away from the total sum. I know the inequality of [integral from n+1 to infinity of An] < [Total sum - Partial sum] < [integral from n to infinity of An].

Obviously, I cannot take an integral of a factorial at this level (Single variable calculus).

Is there another way to estimate the total sum of the series? Any help would be appreciated, thank you.

 

[Ishuda]

"Find the 5th partial sum, and using an error formula, estimate the error of this sum with respect to the sum of the series."

The summation from n=1 to infinity of [(-1)^n]/n!

I evaluated the partial sum through a calculator, my answer was -19/30, or -.63333333


I'm having difficulty estimating how far this partial sum is away from the total sum. I know the inequality of [integral from n+1 to infinity of An] < [Total sum - Partial sum] < [integral from n to infinity of An].

Obviously, I cannot take an integral of a factorial at this level (Single variable calculus).

Is there another way to estimate the total sum of the series? Any help would be appreciated, thank you.

Alternating series test, for example
http://tutorial.math.lamar.edu/Classes/CalcII/AlternatingSeries.aspx

 

[marek]

Alternating series test, for example
http://tutorial.math.lamar.edu/Classes/CalcII/AlternatingSeries.aspx

I have already determined that the series converges absolutely with the ratio test.

How do I continue from here?

 

[marek]

I would like to ask if this is a proper solution:

If I assumed that the partial sum from n=1 to 50 is fairly close to the total sum, could I subtract the partial sum of n=50 from the partial sum of n=5? Would this error be relatively close to the actual error?

 

[Ishuda]

I have already determined that the series converges absolutely with the ratio test.

How do I continue from here?

Sorry, miss-spoke. The Alternating series error estimate which goes something like: If all a_n are non-negative and decreasing, then the error in summing the alternating series in stopping at the nth term is no more than the (n+1)st term. That is, the error E is given by
E = |$\displaystyle \Sigma_{m=0}^{m=\infty} (-1)^m a_m - \Sigma_{m=0}^{m=n} (-1)^m a_m$|
= |$\displaystyle (-1)^{n+1} a_{n+1} +(-1)^{n+2} a_{n+2} +(-1)^{n+3} a_{n+3} + ...$|
=|$\displaystyle a_{n+1} - a_{n+2} + a_{n+3} + ...$|
or since a_n is decreasing
E = $\displaystyle a_{n+1} - a_{n+2} + a_{n+3} + ... = a_{n+1} - (a_{n+2} - a_{n+3}) + ... < a_{n+1}$

 

[Ishuda]

I would like to ask if this is a proper solution:

If I assumed that the partial sum from n=1 to 50 is fairly close to the total sum, could I subtract the partial sum of n=50 from the partial sum of n=5? Would this error be relatively close to the actual error?

Only if your assumption was correct. But what if your assumption was wrong? In this particular case the assumption is correct but it wouldn't always be correct.

BTW: It wouldn't be that you are supposed to recognize the series as e^-1 - 1 and get your error estimate that way, is it?

 

[marek]

Only if your assumption was correct. But what if your assumption was wrong? In this particular case the assumption is correct but it wouldn't always be correct.

BTW: It wouldn't be that you are supposed to recognize the series as e^-1 - 1 and get your error estimate that way, is it?

I knew the value of the series was (e^-1) -1 through wolfram alpha, I'm not sure because I do not have a solution for this particular question.

How did you recognize that the series converged to that particular value?