If $y_n=\int_{1}^{2}\frac{e^{-nt}}{t}\,\mathrm{d}t $ then where does $y_n$ converges?
If $\displaystyle y_n=\int_{1}^{2}\frac{e^{-nt}}{t}\,\mathrm{d}t $, then where does $y_n$ converges?
Try
$$|y_n - y_m|\leq \int_{1}^{2} \frac{|e^{-nt-mt}|}{t} \, \mathrm{d}t \leq \ln 2, $$
which doesn't tend to $0$ as $n$ tend to $\infty$ hence $y_n$ is not convergent and also it is not divergent to $\infty$.
Am I correct. Please help me learn if there is any other way to solve these kind of problems!
$y_n \rightarrow 0$ because $$0 \leq y_n \leq e^{-n}.$$ The right hand side inequality above is true because $e^{-nt}/t \leq e^{-n}$ for each $t \in [1, 2]$.