Convergence of sequence: f(n+1)=f(n)+f(n)2n(n+1)

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Convergence of sequence:  f(n+1)=f(n)+f(n)2n(n+1)

Micah May · Accepted Answer

The sequence is monotone, let g(n)=5−25n, we have f(18)&amp;lt;g(18) and if f(n)&amp;lt;g(n),f(n+1)&amp;lt;f(n)+25n(n+1)&amp;lt;5−25n+25n+1=g(n+1), by induction f(n) is bounded above by 5 therefore converge. However I suggest trying to find a better upper bound g(n) for similar purpose so you don&#039;t need to calculate 18 terms.

Gwendolyn Meyer · Accepted Answer

We know that [f(k)]k=1∞ is monotone increasing. This means that the sequence either diverges to +∞ or converges. In order to determine which option is the case, we do the following trick: First we write  f(k+1)−f(k)=f(k)2k(k+1)≤f(k)f(k+1)k(k+1)  Dividing both sides by f(k)f(k+1) and writing g(k)=1f(k), we have  g(k)−g(k+1)≤1k−1k+1 (1)  This is the key inequality for our argument. One the one hand, summing this for k=1,…,n−1 we get  1−g(n)≤1−1n⇒1n≤g(n)  On the other hand, summing (1) for k=n,n+1,… and denoting g(∞)=limg(n), we have  g(n)−g(∞)≤1n⇒g(n)≤g(∞)+1n  When g(∞)=0, these two inequalities become saturated, hence we obtain g(n)=1n and f(n)=n. But this is absqrt, since this does not satisfy our recurrence relation. Therefore g(∞)&amp;gt;0 and  limn→∞f(n)=1g(∞)&amp;lt;∞

RizerMix · Accepted Answer

Since f(1)=1 and f(n+1)=f(n)+f(n)2n(n+1) (1) we have that f(n) is increasing and inductively we have that 1≤f(n)≤n Furthermore, (1) implies that (1f(n+1)−1n+1)−(1f(n)−1n) =(1n−1n+1)−(1f(n)−n(n+1)n(n+1)f(n)+f(n)2) =1n(n+1)−1n(n+1)+f(n) =f(n)n(n+1)(n(n+1)+f(n)) ≥f(n)n2(n+1)(n+2) ≥1n2(n+1)(n+2) Since ∑n=1∞1n2(n+1)(n+2)=π212−58 we have 1f(n)≥1n+∑k=1n−11n2(n+1)(n+2)≥π212−58 Therefore, f(n)≤242π2−15 Since f(n) is increasing and bounded above, it converges.

Convergence of sequence: f(n+1)=f(n)+\frac{f(n)^2}{n(n+1)}

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