# my teacher marked this wrong on the test

• Dec 10th 2008, 02:35 AM
Legendsn3verdie
my teacher marked this wrong on the test
should i argue with her, or is this just wrong.

i got 0 credit for it, instructions of the test were:
"fore each of the follwoing, determine wheter the infinite series converges or diverges. if it is an alternating series, determine whether it coverges conditionally or absolutely.

http://i176.photobucket.com/albums/w...ntitled4-1.jpg

is that mathmatically wrong somewhere?
• Dec 10th 2008, 02:55 AM
mr fantastic
Quote:

Originally Posted by Legendsn3verdie
should i argue with her, or is this just wrong.

i got 0 credit for it, instructions of the test were:
"fore each of the follwoing, determine wheter the infinite series converges or diverges. if it is an alternating series, determine whether it coverges conditionally or absolutely.

http://i176.photobucket.com/albums/w...ntitled4-1.jpg

is that mathmatically wrong somewhere?

Yes it's wrong. Your teacher is quite correct to give you zero credit for this solution. It shows a fundamental lack of understanding of series algebra.

The result of the nth root test is 2/5, NOT 5*(2/5).

Once you take the 5 out of the summation, your $\displaystyle a_n$ is $\displaystyle \left(\frac{2}{5}\right)^n$.

Once the 5 is taken out, all it does is multiply with the value that the series converges to.
• Dec 10th 2008, 02:58 AM
CaptainBlack
Quote:

Originally Posted by Legendsn3verdie
should i argue with her, or is this just wrong.

i got 0 credit for it, instructions of the test were:
"fore each of the follwoing, determine wheter the infinite series converges or diverges. if it is an alternating series, determine whether it coverges conditionally or absolutely.

http://i176.photobucket.com/albums/w...ntitled4-1.jpg

is that mathmatically wrong somewhere?

The n-th root of the n-th term is $\displaystyle \sqrt[n]{5 (2/5)^n}$ not $\displaystyle 5 \sqrt[n]{(2/5)^n}$

Alternativly you should be applying the n-th root test to the series $\displaystyle \sum_{n=1}^{\infty}(2/5)^n$

CB
• Dec 10th 2008, 03:06 AM
Legendsn3verdie
hm yah ok i guess i was completly wrong thanks for helpin me understand my error.