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convergence of this SEQUENCE
- Thread starter iDoof
- Start date
Does this look familiar?:
\(\displaystyle \lim_{n\to\infty}\(1+\frac{1}{n})^{n}=e\)
You could start by letting, say, \(\displaystyle t=\frac{3}{n}\)
Then \(\displaystyle n=\frac{3}{t}\)
\(\displaystyle \lim_{n\to\infty}(1+\frac{3}{n})^{n}=\lim_{n\to\0}(1+t)^{\frac{3}{t}}\)
\(\displaystyle =\lim_{n\to\0}[(1+t)^{\frac{1}{t}}]^{3}=e^{3}\)
\(\displaystyle (e^{3})^{4}=e^{12}\)
\(\displaystyle \lim_{n\to\infty}\(1+\frac{1}{n})^{n}=e\)
You could start by letting, say, \(\displaystyle t=\frac{3}{n}\)
Then \(\displaystyle n=\frac{3}{t}\)
\(\displaystyle \lim_{n\to\infty}(1+\frac{3}{n})^{n}=\lim_{n\to\0}(1+t)^{\frac{3}{t}}\)
\(\displaystyle =\lim_{n\to\0}[(1+t)^{\frac{1}{t}}]^{3}=e^{3}\)
\(\displaystyle (e^{3})^{4}=e^{12}\)
Hello, iDoof!
We have: .\(\displaystyle \left(1\,+\,\frac{3}{n}\right)^{4n}\)
. . Rewrite the fraction: .\(\displaystyle \frac{3}{n}\:=\:\frac{1}{\frac{n}{3}}\)
. . Multiply the exponent by \(\displaystyle \frac{3}{3}:\;\;4n\cdot\frac{3}{3}\:=\:\left(\frac{n}{3}\right)\cdot12\)
Then we have: .\(\displaystyle \left(1\,+\,\frac{1}{\frac{n}{3}}\right)^{(\frac{n}{3})(12)}\)
Let \(\displaystyle z\,=\,\frac{n}{3}\). . Note that, if \(\displaystyle n\to\infty\), then \(\displaystyle z\to\infty\).
. . The limit becomes: . \(\displaystyle \L\lim_{z\to\infty}\left[\left(1 + \frac{1}{z}\right)^z\right]^{12}\;=\;\left[\lim_{z\to\infty}\left(1\,+\,\frac{1}{z}\right)^z\right]^{12}\;=\;e^{12}\)
\(\displaystyle \L -\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\)
Ha! . . . we had the same idea, galactus!
(You just dropped the "4".)
Are you familiar with the definition of e? . \(\displaystyle \L\lim_{z\to\infty}\left(1\,+\,\frac{1}{z}\right)^z\;=\;e\)I have the general term of a sequence: .\(\displaystyle a_n\;=\;(1\,+\,\frac{3}{n})^{4n}\)
Does it converge or diverge?
We have: .\(\displaystyle \left(1\,+\,\frac{3}{n}\right)^{4n}\)
. . Rewrite the fraction: .\(\displaystyle \frac{3}{n}\:=\:\frac{1}{\frac{n}{3}}\)
. . Multiply the exponent by \(\displaystyle \frac{3}{3}:\;\;4n\cdot\frac{3}{3}\:=\:\left(\frac{n}{3}\right)\cdot12\)
Then we have: .\(\displaystyle \left(1\,+\,\frac{1}{\frac{n}{3}}\right)^{(\frac{n}{3})(12)}\)
Let \(\displaystyle z\,=\,\frac{n}{3}\). . Note that, if \(\displaystyle n\to\infty\), then \(\displaystyle z\to\infty\).
. . The limit becomes: . \(\displaystyle \L\lim_{z\to\infty}\left[\left(1 + \frac{1}{z}\right)^z\right]^{12}\;=\;\left[\lim_{z\to\infty}\left(1\,+\,\frac{1}{z}\right)^z\right]^{12}\;=\;e^{12}\)
\(\displaystyle \L -\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\)
Ha! . . . we had the same idea, galactus!
(You just dropped the "4".)