gobathegreat wrote: Sequence $a_n$ is defined by $a_1=\frac{1}{2}$, $a_m=\frac{a_{m-1}}{2m \cdot a_{m-1} + 1}$ for $m>1$. Determine value of $a_1+a_2+...+a_k$ in terms of $k$, where $k$ is positive integer. So $\frac 1{a_m}=\frac 1{a_{m-1}}+2m$ and so $\frac 1{a_m}=m(m+1)$ and $a_m=\frac 1{m(m+1)}$ $=\frac 1m-\frac 1{m+1}$ So $\boxed{\sum_{m=1}^ka_m=1-\frac 1{k+1}}$

gobathegreat wrote: Sequence $a_n$ is defined by $a_1=\frac{1}{2}$, $a_m=\frac{a_{m-1}}{2m \cdot a_{m-1} + 1}$ for $m>1$. Determine value of $a_1+a_2+...+a_k$ in terms of $k$, where $k$ is positive integer. See here : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=521283&p=2936308#p2936308

The answer is $\frac{k}{k+1}$. Claim: We have $a_m=\frac{1}{m(m+1)}$. Proof. We will proceed by induction. Base case $m=1, 2$ works. Now assume our statement is true for $m$. We will prove for $m+1$. Notice \[ a_{m+1}=\frac{a_m}{2(m+1)\cdot a_m+1}=\frac{1}{m(m+1)}\cdot \frac{m}{m+2}=\frac{1}{(m+1)(m+2)}. \square \] We want $T_k=\sum_{i=1}^k \frac{1}{k(k+1)}=\frac{k}{k+1}$. We will also prove this by induction. Base case works. Now \[ T_{k+1}=T_k+\frac{1}{(k+1)(k+2)}=\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}=\frac{k+1}{k+2}. \]