n=1∑∞(−1)nn2+n+1n2 n=1∑∞2nncosnπ n=1∑∞(−1)n5nn2 n=1∑∞2nncosnπ n=1∑∞(−1)nn3+2n n=1∑∞2nncosnπ n=1∑∞(−1)n5nn2
Posted: Thu Jul 14, 2022 4:41 pm
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n=1∑∞(−1)nn2+n+1n2 n=1∑∞2nncosnπ n=1∑∞(−1)n5nn2 n=1∑∞2nncosnπ n=1∑∞(−1)nn3+2n
n=1∑∞2nncosnπ n=1∑∞(−1)n5nn2 n=1∑∞2nncosnπ n=1∑∞(−1)nn3+2n 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∑∞nxn n=2∑∞n5nxn
\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+2n 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!