# Is $\sum_{n=1}^\infty {(a_n+b_n)}^{-\frac{1}{2n}} = \infty$?

If $$\{a_n\}_{n=1}^\infty$$ and $$\{b_n\}_{n=1}^\infty$$ are two sequences of positive real numbers such that $$\sum_{n=1}^\infty {a_n}^{-\frac{1}{2n}} = \infty$$ and $$\sum_{n=1}^\infty {b_n}^{-\frac{1}{2n}} = \infty$$ then is $$\sum_{n=1}^\infty {(a_n+b_n)}^{-\frac{1}{2n}} = \infty$$ ?

I am trying this real-analysis problem but cannot progress because I can't see how to use some useful inequalities. Can someone please help?

• @DavidC.Ullrich Yes, ok I'll mention that. Commented Aug 16, 2022 at 12:49
• @DavidC.Ullrich Please note the negative exponent. Commented Aug 16, 2022 at 12:54
• @ThomasAhle Please note the negative exponent. Commented Aug 16, 2022 at 12:54
• @DavidC.Ullrich Note that the example you considered with $a_n=b_n=1$ diverges because $\lim_{n \to \infty} 2^{-\frac{1}{2n}}=1$ . Commented Aug 16, 2022 at 13:08
• Thanks, Actually I deleted the answer two minutes before you posted this... Commented Aug 16, 2022 at 13:14

It's not true. Define $$a_{2n+1}=b_{2n}=2^{n^2}$$ and $$a_{2n}=b_{2n+1}=1$$.
• My mistake, it needed some more $n$ in the exponent. Commented Aug 16, 2022 at 13:35
• @DavidC.Ullrich isn't the current answer ok with $2^{n^2}$ ? Commented Aug 16, 2022 at 14:07