Finding Marginal pmf I am studying for an exam and have come across this problem:
Let the random variables $X$ and $Y$ have the joint pmf:
$f_{XY}(x,y)={2\over{n(n+1)}}$ for $y=1, . . . , x$; $x=1, . . . , n$
Find the marginal pmf of $f_Y(y)$.
Normally I have no problem finding these, but this seems to have a dependence in the support that wreaks havoc on my finding the marginal distribution.  I tried this, but this leaves $x$ in the marginal distribution's support, so I know it's not right.
$f_Y(y)=\sum_{x=1}^{n}f(x,y)={2n\over{n(n+1)}}={n\over{n(n+1)}}$ for $y=1, . . . , x$
I know this isn't write since the support now contains, $x$.  The solution is shown in my notes as:
$f_Y(y)={2(n-y+1)\over{n(n+1)}}$ for $y=1, . . . , n$
Can someone help me understand how this solution was derived?  
Thank you.
References
Roussas, George G. An introduction to probability and statistical inference. 2nd ed. San Diego, CA: Elsevier/Academic Press, 2015.  (p. 151, exercise 2.6)
 A: The joint pmf is
$$
  f_{X,Y}(x,y) = \frac{2}{n(n+1)} \times I_{\{1,\dots,n\}}(x) \times I_{\{1,\dots,x\}}(y).
$$
Draw a figure to check that
$$
  I_{\{1,\dots,n\}}(x) \times I_{\{1,\dots,x\}}(y) = I_{\{1,\dots,n\}}(y) \times I_{\{y,y+1,\dots,n\}}(x).
$$
Hence,
\begin{align*}
  f_Y(y) &= \sum_{x=1}^n f_{X,Y}(x,y) = \frac{2}{n(n+1)} \times I_{\{1,\dots,n\}}(y) \times \sum_{x=1}^n I_{\{y,y+1,\dots,n\}}(x) \\
  &= \frac{2(n-y+1)}{n(n+1)} \times I_{\{1,\dots,n\}}(y).
\end{align*}
A: If you try to plot a table of joint probabilities of $f_{XY}$ you will notice that only the upper triangular part of the table has non-zero values, and notice also that:
$$P(y=1)=\sum_{i=1}^{n} \frac{2}{n(n+1)}$$
$$P(y=2)=\sum_{i=2}^{n} \frac{2}{n(n+1)}$$
and so on. So, 
$$P(Y=y)=\sum_{i=y}^{n} \frac{2}{n(n+1)}=\frac{2(n-y+1)}{n(n+1)}$$
And $P(Y=y)$ is just the PMF of $Y$
A: I will post an answer to my own question to add to @Zen's, but @Philip Sarwate's comment helped me the most.  I just took an arbitrary $n$, in this case WLOG, I set $n=5$ and created a $x-y$ table as he suggested and filled in the values:

I noticed the general upper triangular pattern. and noticed that as $y$ increased, the number of constant terms summed (see the values in parentheses at in the last column) to form $f_Y(y)$ decreased and followed the pattern ($n-y+1$).  So, the final answer is simply:
$f_Y(y)=(n-y+1){2\over{n(n+1)}}$ for $y=1, 2, ..., n$ and $0$ otherwise.
Thank you again for everyone's assistance.  This really is a great community.
