Find limit

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  • Thread starter SatyaDas
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  • #1
SatyaDas
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I wonder if the limit of the following can be converted into integral or some elegant form as N tends to infinity:
\[ \sum_{n=0}^{N}\frac{a}{2^{n}}\sin^{2}\left(\frac{a}{2^{n}}\right) \]

If we plot or evaluate the value then it does appear that the series converges very fast.

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Answers and Replies

  • #2
SatyaDas
22
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I forgot to mention that the above graph is plotted as function of x and value of N can be changed by using the slider and that shows that the graph stabilizes pretty fast if N is increased.
 
  • #3
Greg
Gold Member
MHB
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\[ \lim_{N\rightarrow\infty}\frac{x}{2^n}\sin^2\left(\frac{x}{2^n}\right)\rightarrow0\times0^2\rightarrow0 \] but as the limit is taken over positive $x$ the limit tends to infinity.
 
  • #4
SatyaDas
22
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\[ \lim_{N\rightarrow\infty}\frac{x}{2^n}\sin^2\left(\frac{x}{2^n}\right)\rightarrow0\times0^2\rightarrow0 \] but as the limit is taken over positive $x$ the limit tends to infinity.
You missed taking the summation into account. The lower case 'n' is the index for summation and the expression is summed till n=N. We need to find the limit of the sum as the upper case 'N' tends to infinity.
And certainly the limit exists and is non zero that is demonstrated by the graph also. You can you the slider in the graph to change the value of N and see that the graph stabilizes pretty fast.
 
Last edited:

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