Compare
to
. Then in applying the LCT, we have
![\displaystyle\lim_{n\to\infty}\left|\frac{\frac1{\sqrt{n^2+1}}}{\frac1n}\right|=\lim_{n\to\infty}\frac n{\sqrt{n^2+1}}=1](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cleft%7C%5Cfrac%7B%5Cfrac1%7B%5Csqrt%7Bn%5E2%2B1%7D%7D%7D%7B%5Cfrac1n%7D%5Cright%7C%3D%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac%20n%7B%5Csqrt%7Bn%5E2%2B1%7D%7D%3D1)
Because this limit is finite, both
![\displaystyle\sum_{n=1}^\infty\frac1{\sqrt{n^2+1}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac1%7B%5Csqrt%7Bn%5E2%2B1%7D%7D)
and
![\displaystyle\sum_{n=1}^\infty\frac1n](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac1n)
behave the same way. The second series diverges, so
![\displaystyle\sum_{n=0}^\infty\frac1{\sqrt{n^2+1}}=1+\sum_{n=1}^\infty\frac1n](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bn%3D0%7D%5E%5Cinfty%5Cfrac1%7B%5Csqrt%7Bn%5E2%2B1%7D%7D%3D1%2B%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac1n)
is divergent.
Answer:
x = 5, y = 4
Step-by-step explanation:
x + 4y = 21
4x - 16 = y
x + 4(4x - 16) = 21
x + (16x - 64) = 21
(17x - 64) = 21
17x = 85
x = 85/17
x = 5
___________________
x + 4y = 21
5 + 4y = 21
4y + 5 = 21
4y = 16
y = 16/4
y = 4
___________________
Answer:
Step-by-step explanation:
x^2 = 64
√x² = √64
x = 8, -8
answer is B
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