Calculating Limit: \lambda^t/k with 0<\lambda<1

[phonic]

Dear members,

I am calculating the following limit:

<br /> \lim_{t\rightarrow \infty} \sum_{k=1}^{t-2} \frac{\lambda^{t-k-1}}{k}<br />
where
<br /> 0 &lt; \lambda &lt;1<br />

Does anybody know how to do it? Thanks a lot!

 

[rachmaninoff]

A hint:

0\leq \lim_{t\rightarrow \infty} \sum_{k=1}^{t-2} \frac{\lambda^{t-k-1}}{k}=\lim_{t\rightarrow \infty} \lambda^t \sum_{k=1}^{t-2} \frac{\lambda^{k-1}}{k} \leq \lim_{t\rightarrow \infty} \lambda^t \sum_{k=1}^{\infty} \frac{\lambda^{k-1}}{k}

 

[M98Ranger]

Hello,

To calculate this limit, we can use the fact that as t approaches infinity, the sum will approach a Riemann sum for the function f(x) = \frac{\lambda^{x-1}}{x} from k=1 to infinity. This leads to the integral:

\int_{1}^{\infty} \frac{\lambda^{x-1}}{x} dx

Using integration by parts, we can solve this integral to get the final answer of \frac{-\ln\lambda}{1-\lambda}. Therefore, the limit is equal to \frac{-\ln\lambda}{1-\lambda}. I hope this helps! Let me know if you have any further questions.