Question

36. Determine whether the series shown below converges or diverges. k-1 3 3 k=1

Answer 1

Method:There are different ways of series convergence tesing. First of all one can just find series sum. If the value received is finite number, then the series is Converged else Diverged.

So for finding the Sum of the Series, we will use limits i.e.

k-1 3 lim 3

$\lim _{x\to a}\left[c\cdot f\left(x\right)\right]=c\cdot \lim _{x\to a}f\left(x\right)$

$=3\cdot \lim _{k\to \infty \:}\left(\left(\frac{3}{2}\right)^{k-1}\right)$

$=3\cdot \lim _{k\to \infty \:}\left(\left(\frac{3}{2}\right)^k\left(\frac{3}{2}\right)^{-1}\right)$

$=3\left(\frac{3}{2}\right)^{-1}\cdot \lim _{k\to \infty \:}\left(\left(\frac{3}{2}\right)^k\right)$

$=3\cdot \frac{2}{3}\cdot \lim _{k\to \infty \:}\left(\left(\frac{3}{2}\right)^k\right)$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^x=e^{\ln \left(a^x\right)}=e^{x\cdot \ln \left(a\right)}$

$=3\cdot \frac{2}{3}\cdot \lim _{k\to \infty \:}\left(e^{k\ln \left(\frac{3}{2}\right)}\right)$

$=3\cdot \frac{2}{3}\cdot \infty \:$

$=\infty \:$

Since the Sum we received is not finite, therefore, the Series Diverges.

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