- #1
linuxux
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Hello. This is the presented problem:
Suppose [tex](b_{n})[/tex] is a decreasing satisfying [tex]b_{n}\ge\ 0[/tex]. Show that the series
[tex]\sum^{\infty}_{n=1}b_{n}[/tex]
diverges if the series
[tex]\sum^{\infty}_{n=0}{2^{n}b_{2^{n}}}[/tex]
diverges.
I've already proved that i can create [tex]\sum^{\infty}_{n=0}{2^{n}b_{2^{n}}}[/tex] from [tex]\sum^{\infty}_{n=1}b_{n}[/tex] and that that series is larger, so my first idea is to prove this by finding some kind of contradiction by supposing [tex]\sum^{\infty}_{n=1}b_{n}[/tex] converges and trying to prove that the series [tex]\sum^{\infty}_{n=0}{2^{n}b_{2^{n}}}[/tex] diverges.
my idea is if [tex]\sum^{\infty}_{n=1}b_{n}[/tex] is bounded by [tex]M[/tex], then the worst case scenario for [tex]\sum^{\infty}_{n=0}{2^{n}b_{2^{n}}}\ is\ b_{1}\ +\ ...\ +\ 2^{n}M[/tex] but we know there must be something greater since it is unbounded.
i'm not sure if that reasoning works or not.
thanks for the help.
Suppose [tex](b_{n})[/tex] is a decreasing satisfying [tex]b_{n}\ge\ 0[/tex]. Show that the series
[tex]\sum^{\infty}_{n=1}b_{n}[/tex]
diverges if the series
[tex]\sum^{\infty}_{n=0}{2^{n}b_{2^{n}}}[/tex]
diverges.
I've already proved that i can create [tex]\sum^{\infty}_{n=0}{2^{n}b_{2^{n}}}[/tex] from [tex]\sum^{\infty}_{n=1}b_{n}[/tex] and that that series is larger, so my first idea is to prove this by finding some kind of contradiction by supposing [tex]\sum^{\infty}_{n=1}b_{n}[/tex] converges and trying to prove that the series [tex]\sum^{\infty}_{n=0}{2^{n}b_{2^{n}}}[/tex] diverges.
my idea is if [tex]\sum^{\infty}_{n=1}b_{n}[/tex] is bounded by [tex]M[/tex], then the worst case scenario for [tex]\sum^{\infty}_{n=0}{2^{n}b_{2^{n}}}\ is\ b_{1}\ +\ ...\ +\ 2^{n}M[/tex] but we know there must be something greater since it is unbounded.
i'm not sure if that reasoning works or not.
thanks for the help.
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