using power series of e^x, show that if N= 1+1/2+1/3...1/n then e^N>n+1 hence prove series diverges?

Question

help!

Answer 1

According to the properties of the exponencial function, for every real x we have e^x >= 1+ x, with equality if and only if x = 0. Therefore, for every naturaln, it follows that:

e^1 > (1 +1/1)
e^2 > (1 +1/2)
.
.
e^n > (1 +1/n)

Since all terms involved in such inequalities are positive, if we multiply such inequalties and apply the properties of the exponencial function, we get

e^(1 + 1/2 ....+ 1/n) = e^N > (1+1/1) * (1 +1/2)*....(1 +1/n) On the right hand side, we have n terms and n-1 of them are greater than (1 + 1/n). Therefore, e^N > n * (1+ 1/n) = n +1, proving your inequalty. Taking natural logarithms, it follows that N > ln( n+1) for n =1,2,3... Snce ln(n+1) -> oo when n -> oo, it follows N -> oo when n -> oo, so proving the harmonic series diverges.

Answer 2

Genesis a million is the account of the introduction of the universe and life on planet earth because it got here about in chronological series. Genesis 2 is only an elevated rationalization of the activities that befell on the top of the sixth introduction day - at the same time as God created human beings. Genesis one delivers very nearly no information about the introduction of human beings (except the theory that human beings were created in clone of God). For a e book it really is dedicated to the relationship between human beings and God, 4 verses feels like a marginally undesirable rationalization for the introduction of God's preeminent creature. it really is because Genesis one became not in any respect meant to face apart from Genesis 2 and three. Genesis 2 describes God's education of a particular area on the earth (Eden) for habitation by skill of the first human beings. portion of the confusion effects from our English translations, which use the time period "earth" at the same time as the Hebrew would better suited be translated "land."

Answer 3

N= 1+1/2+1/3...1/n
since
1+1/2+1/3...1/n >
{1+1/2 +1/4+1/8 +1/6+1/32..........n} terms
Since RH is in G.P. whose first term is 1 and common ratio is (1/2)
Sn=1{1-0.5^n)/(1-0.5)
=2{1-(1/2)^n}=2-(1/2)^(n-1)
1+1/2+1/3...1/n>2-(1/2)^(n-1)
Therefore
e^(1+1/2+1/3...1/n)>e^{2-(1/2)^(n-1)}
when n tends to infinity then RH=e^2
Therefore it is divergent series
e^N>