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- Thread starter jeca86
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Hello, jeca86!
(a) In the first interpretation, we have: \(\displaystyle \L\,\sum^{\infty}_{n=1}\left[4\,+\,\left(\frac{3}{2}\right)^n\right] \;=\;\sum^{\infty}_{n=1}4 \,+\,\sum^{\infty}_{n=1}\left(\frac{3}{2}\right)^n\)
The first sum is: \(\displaystyle \,4\,+\,4\,+\,4\,+\,4\,+\,\cdots\) which obviously diverges.
(b) In the second version, we have: \(\displaystyle \L\,\sum^{\infty}_{n=1}\left[\frac{4}{2^n}\,+\,\left(\frac{3^n}{2^n}\right)\right] \;=\;\sum^{\infty}_{n=1}\frac{4}{2^n}\,+\,\sum^{\infty}_{n=1}\left(\frac{3}{2}\right)^n\)
The first sum is a geometric series with first term \(\displaystyle a\,=\,\frac{4}{2}\) and common ratio \(\displaystyle r\,=\,\frac{1}{2}\)
\(\displaystyle \;\;\)Since \(\displaystyle r\,<\,1\), this sum converges.
But the second sum has first term \(\displaystyle a\,=\,\frac{3}{2}\) and common ratio \(\displaystyle r\,=\,\frac{3}{2}\)
\(\displaystyle \;\;\)Since \(\displaystyle r\,>\,1\), this sum diverges.
Therefore, the entire sum diverges.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
If it was the second version, a comparison test would have sufficed.
Since \(\displaystyle \L\,\frac{4\,+\,3^n}{2^n}\:>\:\frac{3^n}{2^n}\;\) and \(\displaystyle \L\sum^{\infty}_{n=1}\left(\frac{3}{2}\right)^2\) diverges,
\(\displaystyle \;\;\)then \(\displaystyle \L\,\sum^{\infty}_{n=1}\frac{4\,+\,3^n}{2^n}\,\) also diverges.
Either way, it diverges . . .Convergent or divergent?
\(\displaystyle \L\sum^{\infty}_{n=1}\left[4 \,+\,\frac{3^n}{2^n}\right]\;\) . . . or is it: \(\displaystyle \L\,\sum^{\infty}_{n=1}\frac{4 + 3^n}{2^n}\)
(a) In the first interpretation, we have: \(\displaystyle \L\,\sum^{\infty}_{n=1}\left[4\,+\,\left(\frac{3}{2}\right)^n\right] \;=\;\sum^{\infty}_{n=1}4 \,+\,\sum^{\infty}_{n=1}\left(\frac{3}{2}\right)^n\)
The first sum is: \(\displaystyle \,4\,+\,4\,+\,4\,+\,4\,+\,\cdots\) which obviously diverges.
(b) In the second version, we have: \(\displaystyle \L\,\sum^{\infty}_{n=1}\left[\frac{4}{2^n}\,+\,\left(\frac{3^n}{2^n}\right)\right] \;=\;\sum^{\infty}_{n=1}\frac{4}{2^n}\,+\,\sum^{\infty}_{n=1}\left(\frac{3}{2}\right)^n\)
The first sum is a geometric series with first term \(\displaystyle a\,=\,\frac{4}{2}\) and common ratio \(\displaystyle r\,=\,\frac{1}{2}\)
\(\displaystyle \;\;\)Since \(\displaystyle r\,<\,1\), this sum converges.
But the second sum has first term \(\displaystyle a\,=\,\frac{3}{2}\) and common ratio \(\displaystyle r\,=\,\frac{3}{2}\)
\(\displaystyle \;\;\)Since \(\displaystyle r\,>\,1\), this sum diverges.
Therefore, the entire sum diverges.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
If it was the second version, a comparison test would have sufficed.
Since \(\displaystyle \L\,\frac{4\,+\,3^n}{2^n}\:>\:\frac{3^n}{2^n}\;\) and \(\displaystyle \L\sum^{\infty}_{n=1}\left(\frac{3}{2}\right)^2\) diverges,
\(\displaystyle \;\;\)then \(\displaystyle \L\,\sum^{\infty}_{n=1}\frac{4\,+\,3^n}{2^n}\,\) also diverges.