In order to present $e^x+y=e^xe^y$ by making use of Exponential Series, I obtained the following:

$$e^xe^y=Big(sum_n=0^inftyx^n over n!Big)cdot Big(sum_n=0^inftyy^n over n!Big)=sum_n=0^inftysum_k=0^nx^ky^n over k!n!$$

But, where must I go beside get $e^x+y=sum_n=0^infty(x+y)^n over n!$.

Thanks in advancement.


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You must have actually gotten$$sum_n=0^infty sum_k=0^n fracx^kk!cdotfracy^n-k(n-k)!.$$After that, you can write$$sum_n=0^infty sum_k=0^n frac1n! cdot fracn!k!(n-k)! x^k y^n-k.$$Because the element $dfrac1n!$ does not depfinish on $k$, you have the right to pull it out:$$sum_n=0^infty left(frac1n! sum_k=0^n fracn!k!(n-k)! x^k y^n-k ight).$$Then you have$$sum_n=0^infty frac1n! (x+y)^n.$$

How does $displaystyleleft(sum_n=0^infty b_n ight) left(sum_m=0^infty c_m ight)$ become $displaystylesum_n=0^infty sum_m=0^infty (b_n c_m)$?

And how does $displaystylesum_n=0^infty sum_m=0^infty a_n,m$ come to be $displaystylesum_n=0^infty sum_k=0^n a_k,n-k$?

In the first sum above, alert that $sum_m=0^infty c_m$ does not depend on $n$ so it deserve to be puburned inside the various other amount and also become$$sum_n=0^infty left( b_n sum_m=0^infty c_m ight).$$Then the element $b_n$ does not depend on $m$, so that expression becomes$$sum_n=0^infty sum_m=0^infty (b_n c_m).$$That answers the initially bolded question over.

Next think about the array$$eginarrayccccccccca_0,0 & a_0,1 & a_0,2 & a_0,3 & cdots \a_1,0 & a_1,1 & a_1,2 & a_1,3 & cdots \a_2,0 & a_2,1 & a_2,2 & a_2,3 & cdots \a_3,0 & a_3,1 & a_3,2 & a_3,3 & cdots \vdots & vdots & vdots & vdotsendarray$$The amount $displaystylesum_n=0^infty sum_k=0^n a_k,n-k$ runs down diagonals:$$eginarrayccccccccc& & & & & & & & n=3 \& & & & & & & swarrow \a_0,0 & & a_0,1 & & a_0,2 & & a_0,3 & & cdots \ \& & & & & swarrow \a_1,0 & & a_1,1 & & a_1,2 & & a_1,3 & & cdots \ \& & & swarrow \a_2,0 & & a_2,1 & & a_2,2 & & a_2,3 & & cdots \& swarrow \a_3,0 & & a_3,1 & & a_3,2 & & a_3,3 & & cdots \vdots & & vdots & & vdots & & vdotsendarray$$