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Using one of the following tests: (The Ratio Test / Comparison Test/ Root Test/ Cauchy ratio test / Raabe’s Test) to demonstrates that $\displaystyle\sum_{n=0}^{\infty}(n+2)p^n$ converges if $|p|<1$?

I tired the ratio test

put $a_{n}=(n+2)p^n$ so that $\dfrac{a_{n+1}}{a_n}=\dfrac{(n+3)p^{n+1}}{(n+2)p^{n}}=\dfrac{(n+3)p}{n+2}$ which gives $$R=\lim_{n\to \infty}\mbox{sup}|\dfrac{a_{n+1}}{a_n}|=\lim_{n\to \infty}|\dfrac{(n+3)p}{n+2}|=\lim_{n\to \infty}|\dfrac{np}{n}|=\lim_{n\to \infty}|p|=|p|$$ @Gary @Kavi Rama Murthy thanks for your comments please Does anyone know other test to show that the series converges?

Don freecss
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    You forgot an absolute value sign. You should conclude that $R=|p|$. – Kavi Rama Murthy Oct 05 '21 at 08:10
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    Taking into account the above comment, continue as $$ \mathop {\lim }\limits_{n \to + \infty } \frac{{(n + 3)\left| p \right|}}{{n + 2}} = \left| p \right| \cdot \mathop {\lim }\limits_{n \to + \infty } \frac{{n + 3}}{{n + 2}} = \ldots $$ – Gary Oct 05 '21 at 08:11
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    You can try the root test as well. – Gary Oct 05 '21 at 08:17
  • @Gary It doesn't work for me. I tried it – Don freecss Oct 05 '21 at 08:40
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    Why? $$ \mathop {\lim }\limits_{n \to + \infty } \sqrt[n]{{(n + 2)\left| p \right|^n }} = \left| p \right|\mathop {\lim }\limits_{n \to + \infty } \sqrt[n]{{n + 2}} = \left| p \right|. $$ – Gary Oct 05 '21 at 09:00
  • Thank you, $\lim_{n\to+\infty}\sqrt[n]{{(n + 2)}}=\lim_{n\to+\infty}{{(n + 2)^{\dfrac{1}{n}}}}="(+\infty)^{0}"=1$ true ? – Don freecss Oct 05 '21 at 09:29
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    $$ \lim \sqrt[n]{{n + 2}} = \lim \sqrt[n]{n}\lim \sqrt[n]{{1 + \frac{2}{n}}} = \lim \sqrt[n]{n} $$ and the last limit is classic, see, e.g., https://math.stackexchange.com/questions/115822/how-to-show-that-lim-n-to-infty-n-frac1n-1 – Gary Oct 05 '21 at 10:38
  • So my own way is false. Thank you so much. – Don freecss Oct 05 '21 at 10:52
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    Yes because for example $\lim \sqrt[n]{{n!}} = + \infty$ whereas your way would give $1$. – Gary Oct 05 '21 at 10:55
  • Thank you so much – Don freecss Oct 05 '21 at 10:58

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