Euler's radicalism
Find the value of
e e ⋅ 4 e 3 e ⋅ 6 e 5 e ⋯
The answer is 2.
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2 solutions
Your, LaTeX needs to be corrected man :).
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Found the extra bracket causing the trouble haha
@Adhiraj Dutta , you can use \dfrac \pi 3 3 π for proper size fractions. Similarly for \dbinom mn ( n m ) , while \binom mn ( n m ) . You can also use \displaystyle \frac 1{20}, \binom {20}5, \int \frac \pi 2^\infty, \sum {k=1}^\infty, \lim_{n \to \infty} 2 0 1 , ( 5 2 0 ) , ∫ 2 π ∞ , k = 1 ∑ ∞ , n → ∞ lim . Yes braces { } are optional for single character. We need backslash in front for all function including \ln 2 ln 2 , note that ln is not italic (slanted) and there is a space between ln and 2. All functions including \sin \cos \tan \sec \csc \max \min \gcd ...
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I'm weak with Latex, so I refrain from posting solutions for most problems. Thank you for your advice.
Let , P=(e/e^1/2)(e^1/3/e^1/4)(e^1/5/e^1/6).... Then taking natural log of both sides yields, ln(P)=(1-1/2)+(1/3-1/4)+.... Recognizing the sum is simply ln(2),we the exponential as our result, P=2.
e e ⋅ 4 e 3 e ⋅ 6 e 5 e … can be rewritten as e 1 − 2 1 + 3 1 − 4 1 + 5 1 + …
Now, we know that
ln ( 1 + x ) = x − 2 x 2 + 3 x 3 − 4 x 4 + 5 x 5 + … , f o r − 1 ≤ x ≤ 1
Put x = 1 -
1 − 2 1 + 3 1 − 4 1 + 5 1 + ⋯ = ln 2
Hence now
\[\begin{align*}e^{\displaystyle 1 - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{5} + \dots} &= e^{\ln2} \\
&= \boxed{2} \end{align*}\]