Infinite series challenging problem Let $\sum_{k=1} ^\infty a_k$ be a convergent series of positive terms and let $t_n = \sum_{k=n} ^\infty a_k$ for each integer n. 
I'm trying to prove that $\sum_{k=m} ^n {a_k \over t_k} > 1 - {t_n \over t_m}$ for all positive integers $n$ and $m$ with $m<n$.
We can deduce that ${t_n \over t_m} = {\sum_{k=n}^\infty a_k \over  \sum_{k=m}^\infty a_k} = {\sum_{k=n}^\infty a_k \over  \sum_{k=m}^{n-1} a_k + \sum_{k=n}^\infty a_k}$.
Then, we have  $1-{t_n \over t_m} = {\sum_{k=m}^{n-1} a_k + \sum_{k=n}^\infty a_k \over \sum_{k=m}^{n-1} a_k + \sum_{k=n}^\infty a_k} -  {\sum_{k=n}^\infty a_k \over  \sum_{k=m}^{n-1} a_k + \sum_{k=n}^\infty a_k} = {\sum_{k=m}^{n-1} a_k \over \sum_{k=m}^{n-1} a_k + \sum_{k=n}^\infty a_k}$.
It would be great if $\sum_{k=m} ^n {a_k \over t_k} = {\sum_{k=m}^{n} a_k \over \sum_{k=m}^\infty a_k}$ because since $a_k > 0$ for all $k$, then the conclusion follows. However, I am having trouble convincing myself this is a valid step.
Can you help me? Offer any guidance? Thanks!
 A: Your suspicions are well-founded : your “step” is not valid. But you can 
proceed as follows : if you put
$$
\delta_{m,n}=\bigg(\sum_{k=m}^n \frac{a_k}{t_k}\bigg)-
\bigg(1-\frac{t_n}{t_m}\bigg)
$$
then 
$$
\begin{array}{lcl}
\delta_{m,n} &=& \bigg(\sum_{k=m}^n \frac{a_k}{t_k}\bigg)- \frac{a_{m}+\ldots +a_{n-1}}{t_m} \text{ (as you noticed)} \\
&=& \frac{a_m}{t_m}+ \bigg(\sum_{k=m+1}^n \frac{a_k}{t_k}\bigg)- \frac{a_{m}+\ldots +a_{n-1}}{t_m} \\
&=& \bigg(\sum_{k=m+1}^n \frac{a_k}{t_k}\bigg)- \frac{a_{m+1}+\ldots +a_{n-1}}{t_m} \\
&=& \bigg(\sum_{k=m+1}^n \frac{a_k}{t_k}\bigg)- \frac{t_{n-1}-t_m}{a_m+t_{m+1}} \\
&=& \bigg(\sum_{k=m+1}^n \frac{a_k}{t_k}\bigg)- \frac{t_{n-1}-t_m}{a_m+t_{m+1}}+\frac{t_{n-1}-t_m}{t_{m+1}}-
\frac{t_{n-1}-t_m}{t_{m+1}} \\
&=& \bigg(\sum_{k=m+1}^n \frac{a_k}{t_k}\bigg)+ \frac{a_m(t_{n-1}-t_m)}{t_{m+1}(a_m+t_{m+1})}-
\frac{a_{m+1}+\ldots +a_{n-1}}{t_{m+1}} \\
&=& \frac{a_m(t_{n-1}-t_m)}{t_{m+1}(a_m+t_{m+1})} + \bigg(\sum_{k=m+1}^n \frac{a_k}{t_k}\bigg)-
\frac{a_{m+1}+\ldots +a_{n-1}}{t_{m+1}} \\
&=& \frac{a_m(t_{n-1}-t_m)}{t_mt_{m+1}} + \bigg(\sum_{k=m+1}^n \frac{a_k}{t_k}\bigg)-
\bigg(1-\frac{t_n}{t_{m+1}}\bigg) \\
&=& \frac{a_m(t_{n-1}-t_m)}{t_mt_{m+1}} + \delta_{m+1,n} \\
\end{array}
$$
You can then argue by induction on $n-m$, starting with the base case $n-m=1$, where
$$
\delta_{m,n}=\delta_{m,m+1}=\frac{a_{m+1}}{t_{m+1}}
$$
