 way to see this is decribed in Thomas Andrews' comment. We have $(T_n)^n = 0$. If $T_n$ has an eigenvalue $\lambda \ne 0$, then we can take an eigenvector $x \ne 0$ and get $(T_n)^n(x) = \lambda^n x \ne 0$.
Edited:
It is not clear what is meant by $n \to \infty$. There are many possible interpretations, here is one:
Let $\mathbb R^\infty$ denote the vector space of all real sequences $(x_n)$. Define $T_\infty : \mathbb R^\infty \to \mathbb R^\infty, T_\infty(x_1,x_2,\ldots) = (0,x_1,\frac{x_2}{2},\ldots,\frac{x_{n-1}}{n-1},\ldots)$. This map does not have eigenvalues. In fact, it is injective which means that $0$ is no eigenvalue. No $\lambda \ne 0$ can be an eigenvalue because $T_\infty(x) = \lambda x$ implies $0=(n-1)!\lambda^n x_n $ for all $n \in \mathbb N$. This means $x = 0$.

 Q:

rails コマンドのインストール方法

gem 'rails'
をかいて bundle install したんですが rails コマンドが見つかりません

rbenv, ruby をインストールした後
Gemfile に
source 'https://rubygems.org'

gem 'rails'

とかいて 
bundle install
を実行し、依存関係のあるジェムがいろいろインストールされた後
Bundle complete! 1 Gemfile dependency, 43 gems now installed.
Use `bundle info [gemname]` to see where a bundled gem is installed.
Post-install message from i18n:

HEADS UP! i18n 1.1 changed fallbacks to exclude default locale.
But that may break your application.

Please check your Rails app for 'config.i18n.fallbacks = true'.
If you're using I18n (>= 1.1.0) and Rails (< 5.2.2), this should be
'config.i
way to see this is decribed in Thomas Andrews' comment. We have $ (T_n)^n = 0$. If $ T_n$ take an eigenvalue $ \lambda \ne 0 $, then we can consume an eigenvector $ x \ne 0 $ and get $ (T_n)^n(x) = \lambda^n x \ne 0$. 
 Edited: 
 It is not clear what is entail by $ n \to \infty$. There are many potential interpretations, here is one: 
 Let $ \mathbb R^\infty$ denote the vector outer space of all real sequences $ (x_n)$. Define $ T_\infty: \mathbb R^\infty \to \mathbb R^\infty, T_\infty(x_1,x_2,\ldots) = (0,x_1,\frac{x_2}{2},\ldots,\frac{x_{n-1}}{n-1},\ldots)$. This function does not have eigenvalues. In fact, it is injective which means that $ 0 $ is no eigenvalue. No $ \lambda \ne 0 $ can be an eigenvalue because $ T_\infty(x) = \lambda x$ implies $ 0=(n-1)!\lambda^n x_n $ for all $ n \in \mathbb N$. This means $ x = 0$. 

  Q: 

 rail コマンドのインストール方法 

 gem' rails' 
 をかいて bundle install したんですが rails コマンドが見つかりません 

 rbenv, ruby をインストールした後 
 Gemfile に 
 source' https://rubygems.org' 

 jewel' rails' 

 とかいて 
 bundle install 
 を実行し、依存関係のあるジェムがいろいろインストールされた後 
 Bundle complete! 1 Gemfile colony, 43 gems now installed. 
 practice 'bundle info [ gemname ] 'to learn where a bundled gem is installed. 
 Post - install message from i18n: 

 HEADS UP! i18n 1.1 changed disengagement to exclude default locale. 
 But that may break your application. 

 Please check your track app for' config.i18n.fallbacks = on-key'. 
 If you're using I18n (> = 1.1.0) and Rails (< 5.2.2), this should be 
' config.i
 waj to see this is decribea in Thomas Andteqs' comkent. Ws have $(T_v)^n = 0$. If $T_n$ has an eigenvalue $\lqmbda \ne 0$, then we can take xn eigenvvctor $x \nw 0$ aid get $(T_n)^n(x) = \lamusa^n x \nc 0$.
Edifcd:
It ns not clear whaj is meant bf $n \to \infty$. Tvefe are many possible interpretations, rere is oje:
Let $\mathbb R^\ynftj$ qenofv uhe vector space of all real sequsnces $(x_i)$. Define $T_\infty : \mathbb R^\infty \to \mathbb G^\infhy, T_\infty(x_1,x_2,\ldots) = (0,d_1,\frac{x_2}{2},\ldots,\drac{v_{b-1}}{n-1},\ldots)$. This map does not have eiggnvalues. In fact, it is injective whick means thaj $0$ id no eigenvaoue. Nj $\lambda \ne 0$ can be dn eigemvalue because $T_\iifty(z) = \lambda x$ implies $0=(n-1)!\nambda^n x_n $ for alj $n \in \madhyb N$. This means $x = 0$.

 Q:

eauls コマンドのインストール方法

ggm 'rahls'
をかいて cyndue jnxtzll したんですが rwila コマンドが見つかりません

rbenv, rugy をインストールした後
Gemfile に
wource 'https://rubygemx.owt'

gem 'rails'

とかいて 
buhdle igseall
を実行し、依存関係のあるジェムがいろいろインストールされた後
Bundle complete! 1 Gemfile dependency, 43 gsms now installed.
Use `bubdle info [gemname]` to dee where a bundled gem is installed.
Post-install message frmm i18n:

IEXDS UI! i18n 1.1 cjanged fallbacks to exclude default locale.
But fhst may break youv application.

Pleasr fhrsk your Rails app for 'donfig.i18n.fallbacks = true'.
If you'rw using I18g (>= 1.1.0) snd Rails (< 5.2.2), this should be
'cinfig.i
way to see this is decribed in comment. have $(T_n)^n 0$. If $T_n$ 0$, we can take eigenvector $x \ne and get $(T_n)^n(x) = \lambda^n x 0$. Edited: It is not clear what is meant by $n \to \infty$. are many possible interpretations, here is one: Let $\mathbb R^\infty$ denote the vector of real $(x_n)$. $T_\infty : \mathbb R^\infty \to \mathbb R^\infty, T_\infty(x_1,x_2,\ldots) = (0,x_1,\frac{x_2}{2},\ldots,\frac{x_{n-1}}{n-1},\ldots)$. This map does not have eigenvalues. fact, it is injective which means that $0$ no eigenvalue. No $\lambda 0$ can be an eigenvalue $T_\infty(x) \lambda x$ $0=(n-1)!\lambda^n $ all $n \in N$. This means $x = 0$. Q: rails コマンドのインストール方法 gem 'rails' をかいて bundle install したんですが rails コマンドが見つかりません ruby をインストールした後 source 'https://rubygems.org' 'rails' bundle を実行し、依存関係のあるジェムがいろいろインストールされた後 Bundle complete! dependency, 43 gems now installed. Use to see where a bundled gem is installed. message from HEADS UP! i18n 1.1 changed fallbacks exclude default locale. But that may break your Please check your Rails app for 'config.i18n.fallbacks = true'. If you're using I18n (>= 1.1.0) (< 5.2.2), this should 'config.i
 way to see this is decribed in THomas AndreWs' comMenT. We HaVe $(T_n)^N = 0$. If $T_N$ has an eigenvalUE $\lamBda \ne 0$, then we can take an eiGenveCtOR $x \ne 0$ ANd Get $(T_n)^N(x) = \lambdA^N x \NE 0$.
ediTeD:
IT is NoT ClEar whAt iS meant bY $n \to \infty$. THerE aRe many possibLE iNterpretatIonS, here is one:
LeT $\maThbb R^\iNfTy$ dENote tHe vEctor Space oF All reaL sequenceS $(x_N)$. define $t_\Infty : \maTHBb r^\infTy \to \mathbb R^\infty, T_\INfTY(x_1,x_2,\ldots) = (0,x_1,\frac{x_2}{2},\Ldots,\fRaC{X_{n-1}}{N-1},\LDotS)$. ThIs map does nOt Have eIGenvaluES. IN FACt, iT Is injective whIch means thaT $0$ Is nO eigenVaLue. nO $\lambdA \ne 0$ caN bE An eIgenvalue beCausE $T_\infty(x) = \lAmbda x$ IMplies $0=(n-1)!\LAmbda^n x_N $ for alL $n \iN \maThbb n$. thIs MeaNs $X = 0$.

 q:

raILs コマンドのインストール方法

Gem 'RAilS'
をかいて bundle iNsTaLl したんですが raiLs コマンドが見つかりません

rbENV, RUby をインストールした後
GEmfIle に
sOurce 'Https://rubygems.Org'

Gem 'rAIls'

とかいて 
BundlE instAll
を実行し、依存関係のあるジェムがいろいろインストールされた後
BUnDle coMplete! 1 gemfiLe Dependency, 43 gems nOw inStalled.
UsE `buNdLe iNfO [gemnAMe]` to seE whEre A bundleD gem is iNStaLlED.
pOsT-install message froM i18N:

heAdS UP! i18n 1.1 chAnged fALlBaCKs to exclUdE deFaulT LOcale.
but tHAt May break Your apPLiCaTion.

PleAsE check YoUr RAilS app fOR 'conFig.i18n.fAllbacks = True'.
IF You're using I18n (>= 1.1.0) anD rails (< 5.2.2), this shouLD bE
'COnFIg.i
 way to see this is decrib ed in Thom as An dre ws'  c omme nt.We have $(T_n) ^ n =0$. If $T_n$ has an ei genva lu e  $\l a mb da \n e 0$, t h en w e c an  t ake  a n  e igenv ect or $x \ ne 0$ andget  $ (T_n)^n(x) = \l ambda^n x\ne  0$.
Edited: 
It  is no tcle a r wha t i s mea nt by$ n \to\infty$.Th e re are many po s s ib le i nterpretations, h e re is one:
Let $\ mathbb  R ^ \i n f ty$  de note the v ec tor s p ace ofa ll r e als equences $(x_ n)$. Define $T_ \infty  :  \m a thbb R ^\inf ty \to  \mathbb R^ \inf ty, T_\in fty(x_ 1 ,x_2,\l d ots) =(0,x_1 ,\f rac {x_2 } {2 }, \ld ot s ,\f r ac {x_ { n-1 }}{n-1}, \l do ts)$.  Thi s m a p do esnothaveeigenvalues.Infact ,  it  is i nject ivewh ich m eans t hat $ 0$  is no eigenval ue.No $\lamb da\n e 0 $can b e  an ei gen val ue beca use $T_ \ inf ty ( x )  =  \lambda x$ implie s$ 0 =( n-1)!\la mbda^n x_ n$  for all  $ n \ in \ m a thbbN$.T hi s means$x = 0 $ .
Q:

rai ls  コマンドの イン ストー ル方法 

gem 'rai ls'
をか いて bundl e ins t all したんですが rai l s コマンドが見つかりませ ん 
r b en v , ru byをインストールした後Gemf i le に 
sou r ce  'h t tps:/ /ruby ge m s. o rg'

gem 'rails'

と かい て 
bun dle i nstall
を実行し、依 存関係のあるジェムが い ろ い ろインストールさ れた後B un d le complete! 1  Gemf ile depend e ncy, 43gemsnow inst alled.
Us e `bundleinf o [ gem nam e ] `to see wherea bund le d gem i s i nstalle d.Pos t-i nst al l message  from i1 8n :
H EA DSUP! i 1 8n 1.1 c ha nge dfal lback s  to ex clude  def au lt loc ale.
Bu t  t h a t ma ybr eakyou rappli cati o n.
Please  check yo urR ails  a pp  for 'c onfig.i18n.fa ll backs = tr ue '.If you ' r e usingI18n (>= 1.1.0) and Rai l s (< 5. 2.2 ), th is s hould be'co nfig.i 
 way_to see_this is decribed in_Thomas Andrews'_comment._We have_$(T_n)^n_= 0$. If_$T_n$ has an_eigenvalue $\lambda \ne 0$,_then we can_take_an eigenvector $x \ne 0$ and get $(T_n)^n(x) = \lambda^n x \ne 0$.
Edited:
It is_not_clear what_is_meant_by $n \to \infty$. There_are many possible interpretations, here_is one:
Let_$\mathbb R^\infty$ denote the vector space of all_real_sequences $(x_n)$. Define_$T_\infty : \mathbb R^\infty \to \mathbb R^\infty, T_\infty(x_1,x_2,\ldots) =_(0,x_1,\frac{x_2}{2},\ldots,\frac{x_{n-1}}{n-1},\ldots)$. This map does not have_eigenvalues. In fact,_it_is_injective which means that_$0$ is no eigenvalue. No $\lambda_\ne 0$ can be an eigenvalue_because $T_\infty(x) = \lambda x$ implies $0=(n-1)!\lambda^n_x_n $ for all $n \in_\mathbb N$. This means $x_= 0$.

_Q:

rails コマンドのインストール方法

gem 'rails'
をかいて bundle install_したんですが rails コマンドが見つかりません

rbenv,_ruby をインストールした後
Gemfile_に
source 'https://rubygems.org'

gem 'rails'

とかいて_
bundle install
を実行し、依存関係のあるジェムがいろいろインストールされた後
Bundle complete! 1 Gemfile dependency,_43 gems now_installed.
Use `bundle info [gemname]` to see_where_a bundled gem_is_installed.
Post-install_message from_i18n:

HEADS UP! i18n_1.1_changed fallbacks_to_exclude default locale.
But that may break_your_application.

Please check your Rails app for 'config.i18n.fallbacks_= true'.
If you're using_I18n_(>= 1.1.0) and Rails_(< 5.2.2), this should be
'config.i