Bsrgvty

Does the sequence $e^-(n^frac12)(n+1)^100$ converge? If yes what is the limit?

What I tried: Expanding
$$(n+1)^100= 1+^100C_1n+^100C_2n^2+^100C_3n^3+ dots + ^100C_100n^100$$
Multiplying each term by $e^-(n^frac12)$ and taking limits using L'Hospital Rule we get the limit of the sequence 0, but that is a lengthy and I think it is not a proper approach. [Actually applying L'Hospital rule twice to each term gives the preceding term(I checked it upto $[n^3e^-(n^frac12)$] and hence ultimately limit will be 0 because lim($e^-(n^frac12))=0$]

Can anyone please tell me whether this is a correct method to solve the problem and please suggest me a proper method if there is any. Thank you.

The answer is limit of the sequence is 0

Yes, the final limit is zero. Note that as $nto +infty$
$$e^-sqrtn(n+1)^100=expleft(-sqrtnunderbraceleft(1-frac100ln(n+1)sqrtnright)_to 1right)to0$$
because, for example by using L'Hopital,
$$lim_nto +inftyfracln(n+1)sqrtn=0.$$

Consider

$$f(x)=e^-(x^frac12)(x+1)^100$$

and by $x=y^2to infty$

$$e^-(x^frac12)(x+1)^100=frac(y^2+1)^100e^yto 0$$

indeed for any $m€mathbb N$

$$fracy^me^yto 0$$

for which you can refer to the related

• How to prove that exponential grows faster than polynomial?

Set $m^2=n$ and you get the product of a negative exponential by a polynomial of degree $200$. The exponential always wins.

Don't be impressed by this exponent,

$$sqrt[100]e^-m(m^2+1)^100=e^-m/100(m^2+1)=10000,e^-kk^2+e^-k.$$

As $e^-kto 0$, you can finally reduce to

$$e^-jj.$$

If you know expansion of $e^x=1+x+fracx^22!+fracx^33!+...$
$$lim_n to infty e^-(sqrt n)(n+1)^100=\
lim_n to infty frac(n+1)^100e^sqrt n=\
lim_n to infty frac(n+1)^1001+sqrtn+frac(sqrtn)^22!+fracsqrtn^33!+...+fracsqrtn^201201!+....to 0\$$

Let $n=u^2$, then $$L=lim_u rightarrow infty u^200 e^-u (1+frac1u^2)^100 =lim u^200 e^-u= lim_urightarrow inftyfracu^200e^u rightarrow frac00.$$ Apply L'Hospital Rule D. w. r. t. $u$ up and down separately 200 times to get $$lim_u rightarrow infty frac200! u^0e^u=lim _u rightarrow infty 200!~ e^-infty=0.$$

$e^sqrt n geq frac (sqrt n)^201 (201)!$. Can you complete the proof from this? [$frac (201)! (n+1)^100 n^201/2 to 0$].