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Summarize junior high school math problems
So the partial sum of the first n terms of harmonic series satisfies

sn = 1+ 1/2+ 1/3+…+ 1/n & gt; ln( 1+ 1)+ln( 1+ 1/2)+ln( 1+ 1/3)+…+ln( 1+ 1/n)

= LN2+ln(3/2)+ln(4/3)+…+ln[(n+ 1)/n]

= ln[2 * 3/2 * 4/3 *……*(n+ 1)/n]= ln(n+ 1)

because

lim Sn(n→∞)≥lim ln(n+ 1)(n→∞)=+∞

Therefore, the limit of Sn does not exist, and the harmonic series diverges.

But the limit s = lim [1+1/2+1/3+…+1/n-ln (n)] (n →∞) exists, because

sn = 1+ 1/2+ 1/3+…+ 1/n-ln(n)>ln( 1+ 1)+ln( 1+ 1/2)+ln( 1+ 1/3)+…+ln( 1+ 1/n)-ln(n)

= ln(n+ 1)-ln(n)= ln( 1+ 1/n)

because

lim Sn(n→∞)≥lim ln( 1+ 1/n)(n→∞)= 0

So Sn has a lower bound.

but

sn-S(n+ 1)= 1+ 1/2+ 1/3+…+ 1/n-ln(n)-[ 1/2+ 1/3+…+ 1/(n+ 1)-ln(n+ 1)]

= ln(n+ 1)-ln(n)- 1/(n+ 1)= ln( 1+ 1/n)- 1/(n+ 1)>ln( 1+ 1/n)- 1/n & gt; 0

So Sn decreases monotonously. From the limit theorem of monotone bounded sequence, Sn must have a limit, so

S = lim [1+1/2+1/3+…+1/n-ln (n)] (n →∞) exists.

So let this number be γ, and this number is called Euler constant. The approximate value is about 0.57721566490153286060651209. We don't know whether it is rational or irrational at present. In calculus, Euler constant γ has many applications, such as finding the limit of some series, the sum of some convergent series and so on. For example, to find lim [1/(n+1)+1(n+2)+…+1/(n+n)] (n→∞), you can do this:

lim[ 1/(n+ 1)+ 1/(n+2)+…+ 1/(n+n)](n→∞)= lim[ 1/2+ 1/3+…+ 1/(n+n)-ln(n+n)](n→∞)-lim[65438+65438