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Use the Ratio Test to determine whether the series is convergent or divergent.

sum_(n=1)^(oo)(14^(n))/((n+1)7^(2n+1))
Identify 
a_(n).
Evaluate the following limit.

lim_(n rarr oo)|(a_(n+1))/(a_(n))|
Since 
lim_(n rarr oo)|(a_(n+1))/(a_(n))|?vv1,-- Select-- 
✓.

Use the Ratio Test to determine whether the series is convergent or divergent.\newlinen=114n(n+1)72n+1 \sum_{n=1}^{\infty} \frac{14^{n}}{(n+1) 7^{2 n+1}} \newlineIdentify an a_{n} .\newlineEvaluate the following limit.\newlinelimnan+1an \lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right| \newlineSince limnan+1an?1, \lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right| ? \vee 1,-- Select-- \checkmark .

Full solution

Q. Use the Ratio Test to determine whether the series is convergent or divergent.\newlinen=114n(n+1)72n+1 \sum_{n=1}^{\infty} \frac{14^{n}}{(n+1) 7^{2 n+1}} \newlineIdentify an a_{n} .\newlineEvaluate the following limit.\newlinelimnan+1an \lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right| \newlineSince limnan+1an?1, \lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right| ? \vee 1,-- Select-- \checkmark .
  1. Identify General Term: Identify ana_n as the general term of the series.\newlinean=14n(n+1)72n+1a_n = \frac{14^n}{(n+1)7^{2n+1}}
  2. Evaluate Limit of Ratio: Evaluate the limit of the ratio of successive terms. \lim_{n \to \infty}\left|\frac{a_{n+\(1\)}}{a_{n}}\right| = \lim_{n \to \infty}\left|\frac{\left(\(14\)^{n+\(1\)}\right)/\left((n+\(2\))\(7\)^{\(2\)n+\(3\)}\right)}{\left(\(14\)^{n}\right)/\left((n+\(1\))\(7\)^{\(2\)n+\(1\)}\right)}\right|
  3. Simplify Expression: Simplify the expression inside the limit.\(\newline=limn(14n+1)(n+1)72n+1(14n)(n+2)72n+3= \lim_{n \to \infty}\left|\frac{(14^{n+1})(n+1)7^{2n+1}}{(14^{n})(n+2)7^{2n+3}}\right|\newline=limn14(n+1)72(n+2)= \lim_{n \to \infty}\left|\frac{14(n+1)}{7^2(n+2)}\right|\newline=limn14(n+1)49(n+2)= \lim_{n \to \infty}\left|\frac{14(n+1)}{49(n+2)}\right|
  4. Further Simplify Limit: Further simplify the limit expression.\newline=limn2(n+1)7(n+2)= \lim_{n \rightarrow \infty}\left|\frac{2(n+1)}{7(n+2)}\right|
  5. Evaluate Limit: Evaluate the limit as nn approaches infinity.\newline=limn2(n+1)7(n+2)=27= \lim_{n \to \infty}\left|\frac{2(n+1)}{7(n+2)}\right| = \frac{2}{7}

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