if $S_{m}=\frac{m}{n},S_{n}=\frac{n}{m}(n\neq m)$,then $S_{m+n}$ and $4$ which is bigger? let $\{a_{n}\}$ is  arithmetic sequence,and $S_{n}=a_{1}+a_{2}+\cdots+a_{n}$,and such
$$S_{m}=\dfrac{m}{n},S_{n}=\dfrac{n}{m}(n\neq m)$$
then
$S_{n+m}$ and $4$ which is bigger?
My try: With out loss of we let
$n\ge m$
$$S_{n+m}=S_{n}+a_{n+1}+a_{n+2}+\cdots+a_{n+m}$$
maybe this problem have nice methods,Thank you
 A: $\{a_n\}$ is an arithmetic progression, so $S_n=\frac{n(a_1+a_n)}2=\frac nm$. So $m=\frac2{a_1+a_n}$. Since $m$ is an integer, it implies that $a_1+a_n=1$ or $a_1+a_n=2$. Similarly, $n=\frac2{a_1+a_m}$. So $a_1+a_m=1$ or $a_1+a_m=2$.
Since an arithmetic sequence is monotonous and $n\neq m$, we have $a_n\neq a_m$. So WLOG $a_1+a_n=1$, $a_1+a_m=2$. Then $m=2$ and $n=1$.
Now we can see that $a_1=\frac12$, $a_2=\frac32$. So $d=a_2-a_1=1$ and $a_3=\frac52$. Then
$$S_{m+n}=S_3=\frac{3(\frac12+\frac52)}{2}=\frac92>4$$
A: Note this
$$S_{m+n}=\dfrac{(S_{m}-S_{n})(m+n)}{m-n}$$
then
$$S_{m+n}=\dfrac{(m+n)^2}{mn}>4$$
since $m\neq n$
A: If the sequence is given by $a_k=a+kd$ then you should know that $S_n = na+\frac{n(n+1)}{2}d$. So you have the two lienar equations
$$ ma+\frac{m(m+1)}{2}d=\frac mn,\qquad na+\frac{n(n+1)}{2}d=\frac nm$$
which yoiu can solve for $a$ and $d$ (because $m\ne n$) and hence can eliminate $a,d$ from 
$S_{m+n}=a+\frac{(m+n)(m+n+1)}2d$. Once you have that expression, it should nto be difficult to show that it is or is not greater than $4$.
