Here are some problems in calculus which i am not being able to solve:

\(1.\) Evaluate: \[\large{\lim\limits_{n \to \infty}\frac{1}{2n}log\binom{2n}{n}}\]

\(2.\) Consider the following sequence of positive integers regarding \({a_i}\)

\(a_1=1\) and \(a_n=n(a_{n-1}+1)\) for all \(n≥2\).

Calculate: \[\large{\lim\limits_{n \to \infty}(1+\frac{1}{a_1})(1+\frac{1}{a_2})\dots(1+\frac{1}{a_n})}\]

\(3.\)Let \(f:\mathbb{R}\to \mathbb{R}\) be function that is differentiable \(n+1\) times for some positive integer \(n\). The \(i^{th}\) derivative is denoted by \(f^{i}\). Suppose:

\[f(1)=f(0)=f^{1}(0)=f^{2}(0)=\dots=f^{n}(0)=0\]

Prove that: \(\large{f^{n+1}(x)=0}\)

\(4.\)Let \(f\) be a real valued differentiable function on the real line \(\mathbb{R}\) such that \(\large{\lim\limits_{x\to 0}\frac{f(x)}{x^{2}}}\) exists, and is finite. prove that: \(\large{f'(0)=0}\)

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TopNewest2) \(lim_{n \to \infty} ( \dfrac{a_{1} + 1}{a_{1}}\dfrac{a_{2} + 1}{a_{2}}...\dfrac{a_{n} + 1}{a_{n}}\)

\(a_{n-1} - 1 = \dfrac{a_{n}}{n}\)

Thus,

\( lim_{n \to \infty} (\dfrac{a_{2}}{2}\dfrac{a_{3}}{3}....\dfrac{a_{n+1}}{n+1}.\dfrac{1}{a_{1}a_{2}...a_{n}}\)

\( = lim_{n \to \infty} \dfrac{a_{n+1}}{(n+1)!} = lim_{n \to \infty}\dfrac{1+a_{n}}{n!}\)

\( = lim_{n \to \infty} \dfrac{1}{n!} + \dfrac{a_{n}}{n!} = lim_{n \to \infty} \dfrac{1}{n!} + \dfrac{1}{(n-1)!} \dfrac{a_{n -1}}{(n-1)!}\)

\(lim_{n \to \infty} \dfrac{1}{n!} + \dfrac{1}{(n-1)!} + \dfrac{1}{(n-2)!} + .... \dfrac{1}{(1)!} + \dfrac{a_{1}}{(1)!}\)

\( = e \) as \(a_{1} = 1\) – Megh Choksi · 2 years ago

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– Krishna Sharma · 2 years ago

Great :DLog in to reply

– Aritra Jana · 2 years ago

SUPERB! :DLog in to reply

In 1st use limit as a sum

We will first cancel one \(n!\)

Now it becomes

\(\displaystyle lim_{n \to \infty } \frac{1}{n} log\left( \dfrac{(n+1)(n+2)....(n+n)}{1.2.3...n} \right)\)

\(\displaystyle lim_{n \to \infty } \frac{1}{2n} \sum_{r = 1}^{n} log \left( \dfrac{n + r}{r} \right)\)

Changing summation into integration

\(\displaystyle \frac{1}{2} \int_0^{1} log\left( \dfrac{1+x}{x} \right) dx \)

Now this can be solved easily

4th is easy if limit exist then it is \(\dfrac{0}{0}\) form Applying L-hospital \(\dfrac{f'(x)}{2x}\) so this should also be \(\dfrac{0}{0}\) because limit exist hence

\(f'(0) = 0\)

2nd looks interesting I'll try it – Krishna Sharma · 2 years ago

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– Aritra Jana · 2 years ago

for the 4th one.. that is exactly what i thought.. but i was confused that it might be incorrect.. are you sure that method is right?Log in to reply

– Krishna Sharma · 2 years ago

Yes :), what's the answer to 2nd?Log in to reply

– Aritra Jana · 2 years ago

the answer given is \(\LARGE{\boxed{e}}\)Log in to reply

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3: I don't know if this is a valid proof:

\(f^{(n+1)}(x) = 0\)

iff\(f^{(n)}(x) = C_{n}\) where \(C_{i}\) are some arbitrary constants. However, \(f^{(n)}(0) = 0\); because \(f^{(n)}(x) = f^{(n)}(0) = 0\) for all \(x \in \mathbb{R}\), therefore, \(f^{(n)}(x) = C_{n} = 0\).Repeating the same argument for \(n-1\), \(n-2\), and so on, we see that it must be true that \(f(x) = 0\) and hence \(f^{(n+1)}(x) = 0\).

Logically it seems sound:

\[\forall n \in \mathbb{Z}, x \in \mathbb{R} \Bigg(\boxed{f^{(n+1)}(x) = 0 \wedge f^{(n)}(0) = 0} \longleftrightarrow \boxed{f^{(n)}(x) = C_{n} \wedge f^{(n)}(0) = 0} \longleftrightarrow \boxed{f^{(n)}(x) = 0} \Bigg)\] – Jake Lai · 2 years ago

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– Megh Choksi · 2 years ago

Using PMILog in to reply

– Jake Lai · 2 years ago

What's PMI?Log in to reply

– Megh Choksi · 2 years ago

Principle of mathematical inductionLog in to reply

– Jake Lai · 2 years ago

Yeah, it's pretty much induction but in reverse with base case \(n+1\).Log in to reply

@megh choksi @Krishna Sharma @Sandeep Bhardwaj ..any idea as to how to approach the 3rd one? – Aritra Jana · 2 years ago

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– Jake Lai · 2 years ago

I posted a solution just now, and I think it works. It seems too easy though, which is troubling :<Log in to reply