How do you test the series #Sigma n^2/2^n# from n is #[0,oo)# for convergence?

Answer 1

You can test the series ( \sum_{n=0}^{\infty} \frac{n^2}{2^n} ) for convergence using the ratio test or the root test. Both tests involve taking the limit as ( n ) approaches infinity of a certain expression derived from the terms of the series. If the limit is less than 1, the series converges. If it's greater than 1, the series diverges. If the limit equals 1, the test is inconclusive and another test may be needed.

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Answer 2

the series converges

We can apply d'Alembert's ratio test:

Suppose that;

# S=sum_(r=1)^oo a_n \ \ #, and #\ \ L=lim_(n rarr oo) |a_(n+1)/a_n| #

Then

if L < 1 then the series converges absolutely; if L > 1 then the series is divergent; if L = 1 or the limit fails to exist the test is inconclusive.

So our series is;

# S = sum_(n=0)^oo n^2/2^n #

So our test limit is:

# L = lim_(n rarr oo) | ((n+1)^2/2^(n+1)) / {n^2/2^n} | # # \ \ \ = lim_(n rarr oo) | (n+1)^2/2^(n+1) * 2^n/{n^2} | # # \ \ \ = lim_(n rarr oo) | (n+1)^2/(2*2^n) * 2^n/{n^2} | # # \ \ \ = lim_(n rarr oo) | (n+1)^2/2 * 1/{n^2} | # # \ \ \ = lim_(n rarr oo) | 1/2 * ((n+1)/n)^2 | # # \ \ \ = lim_(n rarr oo) | 1/2 * (1+1/n)^2 | # # \ \ \ = | 1/2 * (1+0)^2 | # # \ \ \ = 1/2 #

and we can conclude that the series converges

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Answer 3

This series converges

The easiest shot is the ratio test , so we look at:

#abs((a_(n+1))/(a_n))_(n to oo#

Here that means:

#abs(((n+1)^2/(2^(n+1)))/(n^2/(2^n)))_(n to oo)#
# = abs(((n+1)^2)/(2 n^2))_(n to oo)#
# = abs((1 + 2/n + 1/n^2)/(2 ))_(n to oo) = 1/2#
So #abs((a_(n+1))/(a_n))_(n to oo) < 1# which means that it converges absolutely.
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Answer 4

To test the convergence of the series (\sum_{n=0}^{\infty} \frac{n^2}{2^n}), you can use the Ratio Test.

  1. Take the limit as (n) approaches infinity of the absolute value of the ratio of successive terms: [ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ] where (a_n) represents the (n)th term of the series.

  2. If the limit is less than 1, the series converges absolutely. If the limit is greater than 1 or if it equals infinity, the series diverges. If the limit equals 1, the test is inconclusive.

For the given series, the (n)th term is (\frac{n^2}{2^n}). So, let's calculate the ratio of successive terms: [ \frac{a_{n+1}}{a_n} = \frac{\frac{(n+1)^2}{2^{n+1}}}{\frac{n^2}{2^n}} = \frac{(n+1)^2}{2(n^2)} ]

Taking the limit as (n) approaches infinity: [ \lim_{n \to \infty} \frac{(n+1)^2}{2n^2} = \frac{1}{2} ]

Since the limit is less than 1, by the Ratio Test, the series (\sum_{n=0}^{\infty} \frac{n^2}{2^n}) converges absolutely.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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