MHB Determining whether the sequence converges or diverges as n--> infinity

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The discussion focuses on determining the convergence or divergence of three sequences as n approaches infinity. The first sequence, a_n = 2 + (-1)^n, diverges due to oscillation. The second sequence, a_n = n/e^n, converges to 0 using L'Hôpital's rule. The third sequence, a_n = (1 + 2/n)^n, is debated; while initially thought to diverge, it actually converges to e^2, depending on the definition of e. The conversation highlights the importance of proper justification in evaluating limits.
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Can someone check my solutions?

Do the following sequences $${a_n}$$ converges or diverge as$$ \n\to\infty$$? If a sequence converges find its limit. Justify your answers.

1. $$a_n = 2 +(-1)^n$$

Answer: so can I say that as lim n --> infinity the sequence diverges by oscillation?

2. $$a_n = \frac{n}{e^n}$$

Answer: so: using l'opitals$$\frac{n}{e^n} = \frac{1}{e^n} = \frac{1}{\infty} = 0$$

$$\therefore$$ sequence converges to 0 ?

3. $$a_n = (1 + \frac{2}{n})^n$$

Answer: So using lopital I got $$(1)^n$$ so would that mean that the sequence diverges becase its 1 + 1 + 1 + 1 to infinity...
 
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shamieh said:
Can someone check my solutions?

Do the following sequences $${a_n}$$ converges or diverge as$$ \n\to\infty$$? If a sequence converges find its limit. Justify your answers.

1. $$a_n = 2 +(-1)^n$$

Answer: so can I say that as lim n --> infinity the sequence diverges by oscillation?

2. $$a_n = \frac{n}{e^n}$$

Answer: so: using l'opitals$$\frac{n}{e^n} = \frac{1}{e^n} = \frac{1}{\infty} = 0$$

$$\therefore$$ sequence converges to 0 ?

3. $$a_n = (1 + \frac{2}{n})^n$$

Answer: So using lopital I got $$(1)^n$$ so would that mean that the sequence diverges becase its 1 + 1 + 1 + 1 to infinity...

These are fine provided you do not have to prove from first principles
 
so is my #3 correct? or will it converge to 1 since its a sequence?
 
shamieh said:
so is my #3 correct? or will it converge to 1 since its a sequence?

I fear I've been writing rubbish. The series converges, in fact it is $e^2$. How you prove that depends on your definition of $e$
 
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Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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