What is the Taylor Series Method for Evaluating Infinite Series?

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In summary, The sum from k = 1 to k = infinity of k^2/k! can be evaluated using the trick of "integrating with respect to 1" and simplifying the summand. This results in the sum being equal to 2e. Another method, using the Taylor series for e^x, can also be used to evaluate the sum, by plugging in x = 1.
  • #1
yxgao
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Evaluate this series: Sum from k = 1 to k = infinity k^2/k!

The answer is 2e.

Thanks so much!
 
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  • #2
What have you tried so far?
 
  • #3
I have no idea how to do this problem. Does it require more knowledge than second year college math? I thought it just requires some knowledge about series, but I can't seem to get the answer. It's been over 3 years since I took calculus so I don't remember much about series expansions. I've tried comparing the series to several common expansions but I can't find the right form or find the right starting point. Thanks!
 
  • #4
yxgao,
here's a hint:
[tex]
e = \sum_{k=0}^\infty \frac{1}{k!}
[/tex]
Try to express your sum in terms of this...
 
  • #5
Have you ever heard of a trick called "integrating with respect to 1"?


Anyways, one of the first things I'd do is simplify the summand; something cancels.
 
  • #6
Here's a hint: replace the index k by k+1. You'll be able to cancel k+1 in numerator and denominator. The rest is easy.
 
  • #7
How do you simplify (n+1)/n! summed from n = 0 to n = infinity?
 
  • #8
distribute!
 
  • #9
I don't understand how to proceed from this step. What do you do with the (n+1) in the numerator? What is there to distribute? Can someone post a direct method to the solution?
Thanks.
 
  • #10
[tex]
\frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}
[/tex]
 
  • #11
Ah, ok. Why didn't I see that before? :)
e + e = 2e
Thanks a lot.
 
  • #12
It's a slick trick, I haven't seen this particular sum done this way before. The method I know can be done as follows (it's more complicated, but more powerful):


The taylor series for [itex]e^x[/itex] is:

[tex]e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!}[/tex]

I can insert a [itex]k[/itex] term by differentiating:

[tex]
e^x = \sum_{k=0}^{\infty} \frac{k x^{k-1}}{k!}
[/tex]

I can then multiply by [itex]x[/itex]:

[tex]x e^x = \sum_{k=0}^{\infty} \frac{k x^{k}}{k!}[/tex]

and differentiate again

[tex]
(1 + x)e^x = \sum_{k=0}^{\infty} \frac{k^2 x^{k-1}}{k!}
[/tex]

Then, plug in [itex]x = 1[/itex] to get the sum.
 

1. What is the purpose of evaluating a series?

Evaluating a series allows us to determine whether the series converges or diverges. This information is important in many areas of science, such as calculus and statistics, where series are used to model real-world phenomena.

2. How do you evaluate a series?

The process of evaluating a series involves finding the limit of the sequence of partial sums. This can be done using various methods, such as the ratio test or the integral test.

3. What does it mean if a series converges?

If a series converges, it means that the sequence of partial sums approaches a finite value as the number of terms increases. In other words, the series has a finite sum and is considered "convergent".

4. What does it mean if a series diverges?

If a series diverges, it means that the sequence of partial sums does not approach a finite value as the number of terms increases. In other words, the series does not have a finite sum and is considered "divergent".

5. Why is it important to determine whether a series converges or diverges?

Knowing whether a series converges or diverges is important in many applications, such as calculating probabilities, approximating functions, and solving differential equations. It also allows us to identify patterns and relationships in data and make predictions based on those patterns.

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