Finding range of value of 75th term in recurrence

Asked Jul 03 '21 at 05:07

Active Jul 03 '21 at 05:07

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Let sequence {$a_n$} be defined as below. $a_1=1$, when $n\geq1$, $a_{n+1}=a_n+\frac{1}{a_n}$

$ a_{75} $ lies between: A) (12,15) B) (15,18) C) (11,12) D) None

An easy observation is that $ a_{75} $ = Σ$\frac{1}{a_n}$ for n=$1$ to $74$

Beyond this I have been able to approximate the values to find a trivial convergence(if any) but found that many initial terms had values in $10$th decimal place (for eg $1,0.5,0.34$ etcetera) so it's difficult to approximate.

I have an intuition that we might have some solution by denoting $ a_n $ with separate variables and integration, but wasn't able to simplify any further due to too many variables.

asked Jul 03 '21 at 05:07

Supersusha22

1

Does this answer your question? Solve the sequences inequality Found using Approach0. – Toby Mak Jul 03 '21 at 05:12
Have you made a spreadsheet and computed the terms? Copy down makes it easy. What have you tried? – Ross Millikan Jul 03 '21 at 05:23
Easy to verify with picture of short and tall rectangles, around the graph of $f(x) = (1/x)$ that $$ \sum_{k=2}^{n} \frac{1}{k} < \int_{1}^n \frac{1}{x}~dx = \log(n) < \sum_{k=1}^{n-1} \frac{1}{k}.$$ – user2661923 Jul 03 '21 at 06:00

Finding range of value of 75th term in recurrence

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