Show that n=1 to ∞ ∑(-1)^n+1 5/7n+2 is conditionally convergent.
∑n=1∞(−1)n+157n+2 − Leibnitz series:altering signs;57n+2↓0,n→∞Hence the series converges.The series ∑n=1∞57n+2does not converge by the comparison test:limn→∞57n+21n=limn→∞57+2n=57and ∑n=1∞1n does not convergeHence the series is conditionally convergent\sum_{n=1}^{\infty}{\left( -1 \right) ^{n+1}\frac{5}{7n+2}}\,\,-\,\,Leibnitz\,\,series:\\altering\,\,signs;\\\frac{5}{7n+2}\downarrow 0,n\rightarrow \infty \\Hence\,\,the\,\,series\,\,converges.\\The\,\,series\ \sum_{n=1}^{\infty}{\frac{5}{7n+2}}\\does\,\,not\,\,converge\,\,by\,\,the\,\,comparison\,\,test:\\\underset{n\rightarrow \infty}{\lim}\frac{\frac{5}{7n+2}}{\frac{1}{n}}=\underset{n\rightarrow \infty}{\lim}\frac{5}{7+\frac{2}{n}}=\frac{5}{7}\\and\,\,\sum_{n=1}^{\infty}{\frac{1}{n}}\,\,does\,\,not\,\,converge\\Hence\,\,the\,\,series\,\,is\,\,conditionally\,\,convergent∑n=1∞(−1)n+17n+25−Leibnitzseries:alteringsigns;7n+25↓0,n→∞Hencetheseriesconverges.Theseries ∑n=1∞7n+25doesnotconvergebythecomparisontest:n→∞limn17n+25=n→∞lim7+n25=75and∑n=1∞n1doesnotconvergeHencetheseriesisconditionallyconvergent
#340153 on Dec 2023