Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series. ∑n=1∞3−n

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Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series.  ∑n=1∞3−n

diskusje5 · Accepted Answer

Integral test:  If f(x) is continuous, positive and decreasing function on [k,∞] and f(n)=an then  If ∫k∞f(x)dx is convergent then ∑n=k∞an is also convergent.  If ∫k∞f(x)dx is divergent then ∑n=k∞an is also divergent.  Given that  ∑n=1∞3−n  Here an=3−n  Since an=3−n so functon f(x) is  f(x)=3−x  F(x) is positive and decreasing so integral test is applicable for the given series.  Now evaluate ∫k∞f(x)dx so  ∫1∞3−xdx=[−3−xln⁡3]1∞  =−3−∞ln⁡3+3−1ln⁡3  =0+13ln⁡3  =13ln⁡3  Since ∫1∞3−xdx is convergent. Hence using integral test ∑n=1∞3−n is also convergent.

Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series. sum_{n=1}^infty3^{-n}

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