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im trying to prove to following:
suppose $a_n , b_n $ sequences such that $a_n , b_n>0, a_n ,b_n\rightarrow0$ than if $\sum^{\infty}_{n=1} a_n$ converges and $\sum^{\infty}_{n=1} b_n$ diverges and $c_n = a_n + b_n$ then$\sum^{\infty}_{n=1} c_n$ diverges.

i proved by contradiction and assumed that it converges. then defined $d_n = c_n - a_n$ from my assumption i can say that $\sum^{\infty}_{n=1} d_n $ converges because its a sum of 2 converging series. then $$\sum^{\infty}_{n=1} d_n =\sum^{\infty}_{n=1} c_n -a_n \overset{\ast}{=} \sum^{\infty}_{n=1} a_n + b_n -a_n = \sum^{\infty}_{n=1} b_n$$ in contradicion to the fact that $\sum^{\infty}_{n=1} b_n$ diverges.

but im not really sure about the $\ast$ step. thank you!

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Why so complicated ? You define $d_n = c_n - a_n$. Are you not aware of $d_n=b_n$ ?

Suppose that $\sum^{\infty}_{n=1} c_n$ converges, then $\sum^{\infty}_{n=1} (c_n-a_n)$ converges. This gives the convergence of $\sum^{\infty}_{n=1} b_n.$ A contradiction !

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  • $\begingroup$ You assume that they know that if $\sum x_n$ converges and $\sum y_n$ converges then $\sum x_n+y_n$ converges. My taste is that these questions are equivalent so it's strange to assume one to prove the other $\endgroup$ – Yanko May 14 '20 at 10:35
  • $\begingroup$ @Yanko i can assume that, i just wanted to know that it is correct to subtract $a_n$ from $c_n$ $\endgroup$ – Elad Elmakias May 14 '20 at 10:37
  • $\begingroup$ @EladElmakias Alright then $\endgroup$ – Yanko May 14 '20 at 10:38
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$$\sum_{n=1}^\infty c_n-a_n = \lim_{N\rightarrow\infty} \sum_{n=1}^N c_n - a_n = \lim_{N\rightarrow\infty} \sum_{n=1}^N a_n+b_n-a_n = \sum_{n=1}^\infty a_n+b_n-a_n$$

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$a_n,b_n >0$;

$b_n \lt a_n+b_n=:c_n$;

$\sum b_n$ diverges implies $\sum c_n$ diverges (Comparison test).

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