n=1∑∞​(−1)nn2+n+1n2​ n=1∑∞​2nncosnπ​ n=1∑∞​(−1)n5nn2​ n=1∑∞​2nncosnπ​ n=1∑∞​(−1)nn3+2​n​ n=1∑∞​2nncosnπ​ n=1∑∞​(−1)n5nn2

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n=1∑∞​(−1)nn2+n+1n2​ n=1∑∞​2nncosnπ​ n=1∑∞​(−1)n5nn2​ n=1∑∞​2nncosnπ​ n=1∑∞​(−1)nn3+2​n​
n=1∑∞​2nncosnπ​ n=1∑∞​(−1)n5nn2​ n=1∑∞​2nncosnπ​ n=1∑∞​(−1)nn3+2​n​ n=1∑∞​(−1)nn2+1n2−1​ n=1∑∞​(−1)n−12nn33n​
n=1∑∞​n3+nn4+1​​ n=1∑∞​n3+1n−1​ n=1∑∞​1+2nsin2n​ n=2∑∞​n​−1(−1)n−1​ n=1∑∞​n​xn n=2∑∞​n5n​xn
\begin{tabular}{l} n=1∑∞​2nn2xn \\ n=1∑∞​(−1)nn2+1n2−1​ \\ n=1∑∞​n3+nn4+1​​ \\ n=1∑∞​5nn​(2x−1)n​ \\ \hlinen=1∑∞​5nn​(2x−1)n​ \end{tabular}
n=1∑∞​(−1)nn3+2​n​ Suppose that for the series Σan​ we have n→−lim​∣an​/an+1​∣=2. What is n→∞lim​∣an+1​/an​∣ ? Does the series converge? n=1∑∞​(−1)n−12nn33n​ 1−1⋅32!​+1⋅3⋅53!​−1⋅3⋅5⋅74!​+⋯+(−1)n−11⋅3⋅5⋅⋯⋅(2n−1)n!​+⋯​ n=1∑∞​(−1)n5⋅8⋅11⋯⋯⋅(3n+2)2nn!​