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ratio test question

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Decide whether the following series converges:

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a) Apply ratio test

\\[\\displaystyle\\lim_{n\\to \\infty} \\frac{\\frac{(n+1)^2+2(n+1)+1}{3^{n+1} + 2}}{\\frac{n^2+2n+1}{3^{n} + 2}} =
\\lim_{n\\to \\infty}\\left( \\frac{(n+1)^2+2(n+1)+1}{3^{n+1} + 2}\\cdot\\frac{ 3^{n} + 2}{n^2+2n+1}\\right)
= \\lim_{n\\to \\infty}\\left( \\frac{(n+1)^2+2(n+1)+1}{n^2+2n+1}\\cdot\\frac{ 3^{n} + 2}{3^{n+1 } + 2}\\right)
=\\frac{1}{3}.\\]

Since the limit is less than $1$, the series is convergent.

Apply ratio test

\\[\\displaystyle\\lim_{n\\to \\infty} \\frac{\\frac{(n+1)^2+2(n+1)+1}{(n+1)!}}{\\frac{n^2+2n+1}{n!}} =
\\lim_{n\\to \\infty}\\left( \\frac{(n+1)^2+2(n+1)+1}{(n+1)!}\\cdot\\frac{ n!}{n^2+2n+1}\\right)
= \\lim_{n\\to \\infty}\\left( \\frac{(n+1)^2+2(n+1)+1}{n^2+2n+1}\\cdot\\frac{ n!}{(n+1)!}\\right)
=\\lim_{n\\to \\infty}\\frac{1}{n} = 0.\\]

Since the limit is less than $1$, the series is convergent.

Apply ratio test

\\[\\displaystyle\\lim_{n\\to \\infty} \\frac{\\frac{(2(n+1)+1)}{5^{n+1}((n+1)!)^2}}{\\frac{(2n+1)}{5^{n}(n!)^2}} =
\\lim_{n\\to \\infty}\\left( \\frac{(2(n+1)+1)}{5^{n+1}((n+1)!)^2}\\cdot\\frac{5^{n}(n!)^2}{2n+1 }\\right)
= \\lim_{n\\to \\infty}\\left( \\frac{2(n+1)+1}{2n+1}\\cdot\\frac{ 5^{n}(n!)^2}{5^{n+1}((n+1)!)^2}\\right)
=\\lim_{n\\to \\infty}\\frac{1}{5n^2} = 0.\\]

Since the limit is less than $1$, the series is convergent.

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$\\displaystyle\\sum_{n=1}^\\infty \\frac{n^2+2n+1}{3^n +2}$

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

$\\displaystyle\\sum_{n=1}^\\infty \\frac{(2n+1)}{5^n (n!)^2}$

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