Infinite Summation problem for CalcII project

[NGMarine]

Hello all,

In my CalcII class we have a project and I'm stuck on one of the problems. I'm trying to find the exact value of the following:

?[n=0,?,[(-1)^n*6^(3n+4)]/[10^(n+1)(3n+3)(3n+4)]]

I'm not sure where to do from here or what Series to use. Any help would be greatly appreciated!

Thank you,
-Kenny

 

[galactus]

Is this your series?:

\(\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^{n}\cdot 6^{3n+4}}{10^{n+1}(3n+3)(3n+4)}\)

If so, this diverges.

For instance, if the upper bound is 20, it sums to \(\displaystyle \approx 1.49\times 10^{25}\)

Gets very large, very fast.

But, if it were \(\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^{n}\cdot 6^{3n+4}}{10^{9n^{3}+30n^{2}+33n+12}}\)

We can factor out \(\displaystyle \frac{6^{4}}{10^{12}}=1.296\times 10^{-9}\)

What remains in the series is 1 at n=0. Any subsequent terms are extremely small, and approach 0 very fast. Big O.

So, it would sum to \(\displaystyle \frac{6^{4}}{10^{12}}\).

 

[NGMarine]

Thanks for the quick reply, galactus.

Your first posted series is the correct one, though I appreciate your full answer and the work you put into it.

Here is the entire problem below. The summation that I'd originally posted was copied off of the board and my professor had written it. He'd also written the steps as seen in lines 2-7 below leading up to the summation I'd originally posted. My professor had further calculated (using Maple) that it should add up to .843 at the 32nd iteration.



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This has been a 'hook-up' problem for the entire class. Some have sought the assistance of other professors in the Math department but the assisting professor later found his work to be wrong.

I'm really wanting to understand how to take this integral more than simply solving the problem, though any assistance is greatly appreciated.

Thanks
-Kenny

 

[galactus]

The first series you posted is not the same as the last one. This is why it is important to post the original problem.

\(\displaystyle 6^{3n+4}\neq 3\cdot 2^{3n+4]\)

I assume you want to integrate \(\displaystyle \int_{0}^{2}ln(1+\frac{x^{3}}{10})dx\)

This can be done by using the series for ln(1+t)

\(\displaystyle ln(1+t)=\sum_{k=0}^{\infty}\frac{(-1)^{k}t^{k+1}}{k+1}\)

By letting \(\displaystyle t=\frac{x^{3}}{10}\), we get:

\(\displaystyle \int_{0}^{2}\sum_{k=0}^{\infty}\frac{(-1)^{k}2^{3k+3}}{10^{k+1}(k+1)}\)

Switching the sum and integral sign (this should be justified, but it is OK here) and integrating, we get:

\(\displaystyle \sum_{k=0}^{\infty}\frac{(-1)^{k}2^{3k+4}}{10^{k+1}(3k+4)(k+1)}\)

\(\displaystyle =\frac{8}{5}\sum_{k=0}^{\infty}\frac{(-1)^{k}2^{3k}}{10^{k}(3k+4)(k+1)}\)

Anyway, the sum of this series is around .3322

I ran out of steam and pushed it through my TI.

Which is the solution to said integral. You do not have to use series to evaluate the integral. You can use more elementary methods, such as

parts, partial fractions, etc.

It is rather tedious either way.

\(\displaystyle ln(1+\frac{x^{3}}{10})=ln(10+x^{3})-ln(10)\)

Using parts:

\(\displaystyle \int ln(10+x^{3})dx=xln(10+x^{3})-3x+30\int\frac{1}{x^{3}+10}dx\)

See?. That icky integral with the cube can be rather onerous.