# calculus - interval of convergence

infinity of the summation n=0: ((n+2)/(10^n))*((x-5)^n)

.. my work so far. i used the ratio test

= lim (n-->infinity) | [((n+3)/(10^(n+1)))*((x-5)^(n+1))] / [((n+2)/(10^n))*((x-5)^n)] |
.. now my question is: was it ok for me to add "+1" to "n+2" to become "n+3"?

= lim (n-->infinity) | [((n+3)/(10^(n+1)))*(((x-5)^(n+1))/1)] * [((10^n)/(n+2))*(1/((x-5)^n))] |

= lim (n-->infinity) | [(((n+3)(x-5))/10)*(1/(n+2))] |

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1. what is the answer x-5/3=10

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posted by roger

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