Download Targil 13 – limits

Targil 13 – limits.
sin m  t 
dt , where m and n are natural numbers,
1. Compute lim 
x 0 
tn
x
(the answer may depend on m and n).
2x
Solution. By mean value theorem of integration
2x
sin m  t 
 sin  t   1
 sin  y  
dt

dt





n
n

m
x t
x  t  t
 y 
m
2x
m 2x
t
1
nm
dt , where x ≤ y ≤ 2x.
x
sin  y 
 sin  y  


1
1 . Therefore
,
so

 
x 0
x 0
y
 y 
2x
2x
2x
sin m  t 
1
mn
 1
 1
lim 
dt  lim  nm dt  nm1  lim  m  n   nm1  1 nm1
n
x0
x0
x0
t
t
t
2
x
x
x
x
if n  m  1. Since the integral is positive anyway, it only matters whether
1
lim nm1 tends to zero or to infinity in these cases, so
x 0 x
if n  m  1 then it tends to infinity,
if n  m  1 then it tends to zero,
and if n  m  1then
2x
2x
sin m  t 
1
lim 
dt  lim  dt  lim  ln  2 x   ln x   ln 2
n
x 0 
x 0 
x 0 
t
t
x
x
Remark. This problem is pretty obvious, but it really did appear on IMC.
m
Hence

2. a) Does
1
 k   ln k    ln  ln k  
converge?
k 100

b) Does
 k
k 100
1
 ln k    ln  ln k  
2
converge?
Solution. a) Apply the “harmonic trick” : split the sequence into sub-strips:
strip number n will contain numbers between 2n+1 and 2n+1.
Then all numbers in the strip number n are (up to a bounded factor)
1
and there are 2n elements in this strip.
n
2  n   ln n 
Therefore the sum of that strip is
1
.
n   ln n 

So it is enough to investigate the convergence of the series
1
 n   ln n  .
n ?
Apply the harmonic trick again: cut the sequence into strips, strip number m
will contain n’s between 2m+1 and 2m+1.
1
The elements in strip number m are, up to a bounded factor m
, this strip
2 m
1
has 2m elements, so the sum in the strip is, up to a bounded factor .
m

1
So the sum over all strips behaves like  , and that one diverges (the
m ? m
easiest proof for that is applying harmonic trick again.
b) As in a), apply the harmonic trick twice. Each time you get the series with

1
the same convergence properties. After firs time you get 
, and
2
n

ln
n
n ?
 

after second time you get
1
m
m ?
2
, and that one converges.
3. Let {an} be a sequence defined by a0 = 1 and recursive formula
1 n
ak
an1 

n  1 k 0 n  k  2

a
Find the limit  kk .
k 0 2

Solution. Consider a generating function of that sequence f  x    an x n .
n 0
1
Then the limit of the series that we should compute is simply f   .
2
n
ak
The equation  n  1 an1  
after we multiply both sides by xn and
k 0 n  k  2

 n

x nk
xm 
k

ak x k
sum them up turn to   n  1 an1   ak x

n  k  2 m 0 m  2 k 0
n 0
n 0 k 0

xm
m 0 m  2
f ' x   f  x  

xm
1 x x 2 x3
     ... .
2 3 4 5
m 0 m  2
x 2 x3 x 4 x5
But we know that  ln 1  x   x      ...
2 3 4 5
x  ln 1  x 
Therefore g  x   
.
x2
x  ln 1  x 
df

dx
Hence
f
x2
So, we should try to compute g  x   
ln 1  x 
 ln 1  x  
dx
1
ln f  const     
dx


ln
x




  x  x 1  x  dx 
x
x2
\
  ln x 
ln 1  x 
ln 1  x 
1
1 
1

dx   ln x 
  
 dx 
x
x 1  x 
x
x
1

x


ln 1  x 
1 x
 ln x  ln 1  x  
ln 1  x 
x
x
When x = 0 the right hand side might be seen as a limit (since it comes from
power series) so it is –1 ; f (0) = 1 and its ln is 0, hence left hand side is
equal to const, so const = –1.
 1 x

ln 1  x  
So f  x   exp 1 
x



e
1
 1 
Therefore f    exp 1  ln     e1ln 2 
2
2
 2 

  ln x 
4*. Compute lim
sin  tan  x    tan  sin  x  
.
arcsin  arctan  x    arctan  arcsin  x  
Simple lemma. Suppose we have a function f (x) such that f (0) = 0 and f
has a continuous derivative around 0.
f u   f u 
Suppose also that u, v  0 and always u  v . Then
 f ' 0 .
u v
f u   f u 
Proof. By Lagrange theorem,
 f '  w where w is between u
u v
and v, so when u, v  0 , w also tends to 0. QED of lemma.
x 0
Notice that all functions are analytic.
sin  tan  x    tan  sin  x   at least sometimes, for example when
3 
x  arctan    because, then LHS is negative and RHS is positive.
2 
Since functions are analytic, their points of coincidence are isolated, so at
some neighborhood of 0, except for 0 itself, sin  tan  x    tan  sin  x   , so
we learn from complex functions. So, at some neighborhood of 0 their
inverse functions, arcsin  arctan  x   ,arctan  arcsin  x   also don’t coincide,
hence we can consider that fraction without problems.
From now on we shall say p(x) ~ q(x) to indicate that lim
x 0
p x
 1.
q  x
It is known that x ~ sin x ~ tan x.
Denote y  tan  sin  x   then y ~ x.
lim
x 0
sin  tan  x    tan  sin  x  
arcsin  arctan  x    arctan  arcsin  x  
 lim
x 0
 lim
x 0
 

sin tan arcsin  arctan  y    y

arcsin  arctan  x    arctan  arcsin  x  


 
x  tan  sin  arctan  arcsin  x    
sin tan arcsin  arctan  y    y
The last equality follows from the above lemma, when you apply function
f  x   tan  sin  x   to u  arcsin  arctan  x   , v  arctan arcsin x   , you get

 
If we denote g  y   sin  tan  arcsin  arctan  y     , and its inverse function
g  x   tan  sin  arctan  arcsin  x     , then what we got is
arcsin  arctan  x    arctan  arcsin  x   ~ x  tan sin arctan  arcsin  x   .
1
lim
x 0
g  y  y
x  g 1  x 
From x ~ sin x ~ tan x it follows that x ~ g  x  ~ g 1  x  .
But we have seen before that both numerator and denominator are nonzero,
so g has no stable points in the neighborhood of 0 other than 0.
Hence Taylor series of g has at least 2 nonzero terms, and the first is x:
g  x   x  ax n  ... , where a is an nonzero number which I know nothing
about, and n > 1 is a number which I know almost nothing about, and I don’t
want to (though they can be computed). Then g 1  x   x  ax n  ... .
Now since y ~ x,
ay n  o  y n 
g  y  y
lim
 lim
1
x 0 x  g 1  x 
x 0
 ay n  o  y n 


So, the answer is 1.
5*. {an} is a sequence of positive real numbers such that an + am ≥ an+m for
all m, n.
Prove that the sequence an/n converges.
Proof. Firstly, an ≤ na1 hence the sequence an/n is bounded from above. It is
also bounded from below by 0. So it is bounded. Hence it has liminf (lower
limit) which will be denoted by L.
For each ε > 0, we can find an index k such that ak/k < L + ε.
Then ak < k(L + ε) .
Then amk ≤ mak < mk(L + ε) .
For each natural N, we can write N = mk + r , where r < k (division with
remainder). Hence aN ≤ amk + ar ≤ mak + ra1 < mk(L + ε) + ma1 .
Therefore aN / N < L + ε + ma1 / N < L + 2ε , for sufficiently large N.
So, for any ε > 0, aN / N < L + 2ε, for sufficiently large N.
Since L is liminf, then for sufficiently large N also aN / N > L – 2ε .
Hence the sequence converges to L.