What is the sum of the series from k=1 to k=Infinity for k^2/k!?

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SUMMARY

The sum of the series from k=1 to k=Infinity for k^2/k! is definitively 2e. This conclusion is derived from the relationship between the series and the exponential function, specifically utilizing the Taylor series expansion for e^x. The manipulation of the series shows that k^2/k! can be expressed in terms of the series for e, leading to the final result of 2e.

PREREQUISITES
  • Understanding of Taylor series, specifically e^x = Σ (x^n / n!)
  • Familiarity with factorial notation and its properties
  • Basic knowledge of limits and convergence of series
  • Ability to manipulate algebraic expressions involving series
NEXT STEPS
  • Study the derivation of the Taylor series for e^x in detail
  • Explore the properties of factorials and their growth rates
  • Learn about convergence tests for infinite series
  • Investigate other series representations of exponential functions
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Students in calculus, mathematicians, and anyone interested in series convergence and the properties of exponential functions.

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Homework Statement



Series k=1 to k=Infinity of:

k^2/k!

a) e
b) 2e
c) (1+e)(e-1)
d) e^2
e) Infinity


Homework Equations



e^x = Series from n=1 to n=Infinity of:
x^n / n!

The Attempt at a Solution



I was guessing this would end up being C but the answer is infact B. As far as I can tell 2e is equal to Series 2^k / k!, I'm unsure how this is equal.
 
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You don't need to get to x^n/n!, but to 1/n!, since this series gives e. A start:

<br /> \frac{{k^2 }}{{k!}} = \frac{{kk}}{{k\left( {k - 1} \right)!}} = \frac{k}{{\left( {k - 1} \right)!}}<br />
 

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