​Hi there,

I bump into a weird sequence, and know for a fact the following holds.
(Also, I ran MATLAB simulations to double-check.) This sequence comes up when doing research on computing the reliability of a signal to reach the destination when sent through parallel links and the benefit associated with it. (Details can be too long, which might be helpful to get the solution, but I think those who are very good at dealing with sequences don't need it necessarily.) 

For any positive integers $n$ and a real number $t \in (0, 1)$, the following holds:

$\frac{n}{\frac{1}{n} \sum\_{i=1}^{n} \frac{1}{1-t^i}} + \frac{t}{1-t} \left( 1 - t^n \right) \leq n$.

I'm struggling to prove this analytically, though. A few techniques that I have tried usually aim at showing an upper bound of LHS is less than or equal to RHS:

The fact that $1 - t \leq 1 - t^i \leq 1 - t^n$ holds for all $i = 1, \dots, n$ implies that $\frac{1}{1-t^n} \leq \frac{1}{n} \sum\_{i=1}^{n} \frac{1}{1-t^i} \leq \frac{1}{t}$ holds for all $i = 1, \dots, n$. 

Then, LHS is less than or equal to

$n(1-t^n) + \frac{t}{1-t} (1-t^n)$.

Subtracting RHS with the upper bound above, we get

$\frac{t}{1-t} \left( -1 + n(1-t)t^{n-1} + t^n \right)$.

The terms in the parentheses imply that it is less than zero: consider a binomial distribution with paramters $n$ and $t$.

A few other attempts similar to the above one failed. I also tried to use Bernoulli's inequality $(1+x)^n \geq 1+nx$, but it didn't help. Maybe I applied the techniques sloppily.

I feel like we should either prove it by induction (when $n=1$, LHS = RHS. Assuming LHS $\leq$ RHS for $n$, we could show it also is true for $n+1$.), or we begin with defining a sequence, say $S\_n = \frac{1}{n} \sum\_{i=1}^{n} \frac{1}{1-t^i}$, and show that the forward difference of LHS is less than or equal to the forward difference of RHS (which is 1). 

But, even after hours of effort, I couldn't seem to find a clue. Could anyone help? Thanks!