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Mazyrski [523]
3 years ago
12

Help my friend out?

Mathematics
2 answers:
IrinaVladis [17]3 years ago
5 0

Answer:

13. -2

14. 0

15. 2

16. 0

Step-by-step explanation:

Since the function is originally f(x), you go to wherever the x value is and see what the y value is.

So...

f(2) means go to x = 2 and see that y = -2

posledela3 years ago
3 0
13. -2
14. 0
15. 2
16. 0
hope this helps:)
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I guess the series is

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We have

\displaystyle\lim_{n\to\infty}\left|\frac{\frac{2^{n+1}(n+1)!}{(n+1)^{n+1}}}{\frac{2^nn!}{n^n}}\right|=2\lim_{n\to\infty}\left(\frac n{n+1}\right)^n

Recall that

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In our limit, we have

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On the chance that you meant to write

\displaystyle\sum_{n=1}^\infty\frac{2^n}{n!n^n}

we have

\displaystyle\lim_{n\to\infty}\left|\frac{\frac{2^{n+1}}{(n+1)!(n+1)^{n+1}}}{\frac{2^n}{n!n^n}}\right|=2\lim_{n\to\infty}\frac1{(n+1)^2}\left(\frac n{n+1}\right)^2

=\displaystyle2\left(\lim_{n\to\infty}\frac1{(n+1)^2}\right)\left(\lim_{n\to\infty}\left(\frac n{n+1}\right)^n\right)=2\cdot0\cdot e=0

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